Convert Binary Search Tree to Sorted Doubly Linked List
To convert a Binary Search Tree (BST) to a sorted Circular DoublyLinked List, we’ll need to perform an inorder traversal of the tree and update the left and right pointers accordingly. Since the traversal is inorder (leftrootright), the linked list will be sorted.
Here’s a stepbystep approach:
 If the tree is empty, return
None
.  Traverse the left subtree, which returns the head of the left linked list.
 Update the current node’s left pointer to the tail of the left linked list, and update the tail of the left linked list’s right pointer to the current node.
 Traverse the right subtree, which returns the head of the right linked list.
 Update the current node’s right pointer to the head of the right linked list, and update the head of the right linked list’s left pointer to the current node.
 Link the tail of the right linked list to the head of the entire linked list to make it circular.
 Return the head of the entire linked list, which is the smallest element.
Here’s the code:


This code performs an inorder traversal of the tree and updates the left and right pointers to create the sorted Circular DoublyLinked List. It returns the pointer to the smallest element of the linked list, which is the head. The time complexity is O(n), where n is the number of nodes in the tree, and the space complexity is O(h), where h is the height of the tree (for the recursion stack).
Identifying Problem Isomorphism
“Convert Binary Search Tree to Sorted Doubly Linked List” can be mapped to “Flatten Binary Tree to Linked List”.
Both problems deal with transformations from a tree data structure to a list. “Flatten Binary Tree to Linked List” requires transforming a binary tree into a singly linked list, while “Convert Binary Search Tree to Sorted Doubly Linked List” requires transforming a binary search tree into a doubly linked list.
While the specific transformations required by these two problems are different, the underlying concept of performing a transformation from a tree to a list is common between them. In both cases, a careful traversal of the tree is needed to ensure that the nodes are arranged in the correct order in the resulting list.
“Convert Binary Search Tree to Sorted Doubly Linked List” is more complex because it requires maintaining two links (previous and next) for each node, instead of just one as in “Flatten Binary Tree to Linked List”. Additionally, because the input is a binary search tree, the doubly linked list needs to be sorted, which adds an additional constraint to the problem.
Understanding how to solve “Flatten Binary Tree to Linked List” can provide a useful starting point for solving “Convert Binary Search Tree to Sorted Doubly Linked List”, but additional steps will be necessary to fully solve the problem.
10 Prerequisite LeetCode Problems
For “426. Convert Binary Search Tree to Sorted Doubly Linked List”, the following are a good preparation:
“94. Binary Tree Inorder Traversal”  This problem helps to understand the inorder traversal in a binary search tree which provides sorted order of the nodes, relevant to converting BST to sorted doubly linked list.
“138. Copy List with Random Pointer”  Although this problem deals with a different data structure (linked list), it provides practice on manipulating node pointers, which is directly relevant to the main problem.
“108. Convert Sorted Array to Binary Search Tree”  This problem offers practice on conversion between a sorted structure (array) and a BST, helpful for understanding the sorted nature of BSTs.
“114. Flatten Binary Tree to Linked List”  This problem requires converting a binary tree into a linked list, directly relevant to the main problem of converting BST to a doubly linked list.
“109. Convert Sorted List to Binary Search Tree”  Understanding this problem can help to comprehend the similar properties between sorted list and BST, and how the transformation happens.
“173. Binary Search Tree Iterator”  This problem helps to understand how to iteratively traverse a BST in a sorted order which is useful in the main problem.
“206. Reverse Linked List”  This problem helps to understand how to manipulate linked list node pointers which will be used in the main problem to create a doubly linked list.
“430. Flatten a Multilevel Doubly Linked List”  Understanding how to work with doubly linked lists in this problem can be beneficial for the main problem.
“144. Binary Tree Preorder Traversal”  It helps understanding the structure of a binary tree, although the order is different, the traversal logic is important.
“145. Binary Tree Postorder Traversal”  It helps understanding the structure of a binary tree, and helps to understand the structure of a BST better.
Understanding binary trees, binary search trees, and doubly linked lists, and how to manipulate their structures and traverse them is important to solving the main problem.
Problem Classification
The problem falls under the domain of Data Structures, specifically Binary Search Trees (BST) and Circular DoublyLinked Lists.
What Components
 A Binary Search Tree (BST) given as input.
 Conversion of this BST to a Circular DoublyLinked List.
 Inplace transformation, meaning no additional memory should be used.
 The linked list must be sorted.
 The first and last elements of the list should be connected to form a circle.
 The output should be a pointer to the smallest element in the linked list.
 Data Structure Manipulation: Converting one data structure (BST) to another (Circular DoublyLinked List).
 Sorting Requirement: The elements in the linked list must be sorted.
 Memory Constraint: The transformation should be inplace, indicating a memory constraint.
 Traversal Problem: Implicit in converting a BST to a list is some form of tree traversal.
 Output Specification: The problem asks for a specific pointer (to the smallest element) as output.
This problem combines elements of data structure manipulation, sorting, memory management, traversal, and specific output requirements. It is essentially a conversion task with various constraints and requirements.
Clarification Questions
 What should be the behavior if the input BST is empty? Should the function return a null pointer or some specific value?
 Are the values in the BST guaranteed to be unique?
 Is the input BST guaranteed to be balanced?
 Are there any memory constraints besides the inplace requirement?
 What is the maximum number of nodes the BST can have, if there is a limit?
 Are negative values allowed in the BST nodes?
 Is it required to keep the original BST structure intact after the conversion, or is it acceptable if it changes?
 How should the function indicate the start of the circular doublylinked list? Is it always the smallest element?
 Should the function return a new type of node for the doublylinked list, or should it repurpose the existing tree nodes?
 Are there any runtime performance requirements for this task?
Problem Analysis and Key Insights
InPlace Transformation: The transformation must be done in place, meaning memory efficiency is a key constraint.
DoublyLinked and Circular: The resulting list must be a doublylinked list and circular, meaning the last element must point to the first as its successor and vice versa.
Predecessor and Successor: The left and right pointers in the BST nodes should become the predecessor and successor pointers in the doublylinked list, which aligns with the inorder traversal of a BST.
Unique Values: All values in the tree are unique, simplifying the transformation as there will be no duplicate nodes.
Return Smallest Element: The function must return a pointer to the smallest element, which is also the starting point of the circular doublylinked list.
Constraints: The constraints on the number of nodes and their values are fairly generous, but still defined, which suggests that computational complexity might not be a significant issue but should still be considered.
These insights provide clues about what kind of algorithmic approach should be taken and what edge cases and constraints must be considered.
Problem Boundary
The scope of the problem is welldefined and limited to transforming a given Binary Search Tree (BST) into a sorted, circular, doublylinked list inplace. The task includes:
 Utilizing existing nodes: No new nodes should be created; the transformation must occur in place.
 Handling pointers: The left and right pointers in the BST should be converted into predecessor and successor pointers in the doublylinked list.
 Circular List: The last node must point back to the first node, and the first must point to the last, making the list circular.
 Sorted Order: The doublylinked list must maintain the sorted order of elements.
 Return Node: The solution must return the smallest element in the list, which will serve as the starting point for traversal.
The problem is restricted to these specific tasks and does not extend to things like handling duplicate values, balancing the BST, or considering external storage. It’s a standalone problem focused on data structure manipulation.
To establish the boundary of the problem, you should consider the following constraints and conditions:
 Input Size: The number of nodes in the tree is between 0 and 2000.
 Value Range: Each node’s value is between 1000 and 1000.
 Uniqueness: All tree node values are unique.
 Return Type: The function must return a pointer to the smallest element in the list.
 InPlace Operation: Transformation must occur inplace, meaning no new nodes should be created.
These constraints define the problem’s scope and set the limits within which your solution must operate. Anything outside these boundaries is not part of the problem and should not be considered in the solution.
Distilling the Problem to Its Core Elements
Fundamental Concept: The fundamental concept here is data structure transformation. Specifically, this involves converting a Binary Search Tree (BST) into a circular doublylinked list while maintaining sorted order. It touches on the principles of tree traversal and linked list manipulation.
Simple Description: We have a tree where each node has at most two children. The tree is special in that smaller numbers are on the left and larger numbers are on the right. We want to change this tree into a loop of nodes, still keeping the numbers in order, and do it without creating any new nodes.
Core Problem: Convert a given BST into a sorted, inplace circular doublylinked list.
Key Components:
 Traversing the BST in sorted order.
 Modifying the left and right pointers in each node to serve as predecessor and successor pointers in a doublylinked list.
 Ensuring the list is circular, i.e., tail connects to head.
Minimal Set of Operations:
 InOrder Traversal of BST to visit nodes in sorted order.
 Update pointers during traversal to form a doublylinked list.
 Finally, connect the last node to the first to make it circular.
By understanding these aspects, you can approach solving the problem in a structured manner.
Visual Model of the Problem
Visualizing this problem can be done through diagrams or stepbystep transformations to help you understand the transition from a Binary Search Tree to a Circular DoublyLinked List.
Binary Search Tree (BST) Diagram: First, draw the initial BST based on the problem statement. For example, if the root is
[4,2,5,1,3]
, sketch a tree where4
is the root,2
and5
are its left and right children, and so on.4 / \ 2 5 / \ 1 3
Traversal Path: Draw arrows or lines to indicate the order in which you will traverse the BST. For an inorder traversal, you’d go
1 > 2 > 3 > 4 > 5
.4 / \ 25 / \ 13
DoublyLinked List Transformation: Next to your BST, draw an empty set of nodes to represent the future doublylinked list. As you traverse the BST, populate this list.
DoublyLinked List: 1 <> 2 <> 3 <> 4 <> 5
Circular Connection: Finally, indicate that the list is circular by connecting the last element back to the first.
DoublyLinked List: 1 <> 2 <> 3 <> 4 <> 5 (and 5 connects back to 1)
Pointer Update: Optionally, you could represent the transformation stepbystep, showing how the left and right pointers in each tree node are updated to form the doublylinked list.
By visualizing the problem in this way, you get a concrete picture of what you’re aiming to achieve, making it easier to conceptualize the solution.
Problem Restatement
Convert an existing Binary Search Tree (BST) into a Circular DoublyLinked List without creating new nodes. After the transformation, each node’s left and right pointers should act as predecessor and successor pointers, respectively, in the linked list. The linked list should be circular, meaning the last element connects back to the first. Return a pointer to the smallest element in the linked list. Constraints: The tree will have between 0 and 2000 nodes, and each node value will be unique and range between 1000 and 1000.
Abstract Representation of the Problem
Transform a tree data structure with sorted nodes into a circular doublylinked list in a way that retains the sorted order. The transformation should be inplace, reusing the nodes of the tree. The tree and the list have the same elements; they are just arranged differently. The goal is to return a reference to the smallest element in the list, which also serves as the entry point to the circular list. The number of elements is bounded, and each element has a unique, constrained value.
Terminology
Binary Search Tree (BST): A tree data structure where each node has at most two children, referred to as the left child and the right child. For each node, all elements in the left subtree are less than the node, and all elements in the right subtree are greater.
Circular DoublyLinked List: A linear data structure where each element points to its next element and its previous element, and the last element is connected back to the first, making it circular.
InPlace Transformation: Altering the original data structure to form a new one without using additional memory for storing a separate structure.
Predecessor and Successor: In a sorted list, the predecessor of an element is the element that comes right before it, and the successor is the element that comes right after it. In the circular doublylinked list, the predecessor of the first element is the last element, and the successor of the last element is the first element.
Node: An individual element in the tree or list, containing the value and pointers to other nodes.
Role within the context:
 BST serves as the initial structure that needs to be transformed.
 Circular DoublyLinked List is the final structure we need to achieve.
 InPlace Transformation specifies the constraint that we cannot use additional memory proportional to the input size.
 Predecessor and Successor help define the relationships between nodes in the transformed structure.
 Node is the fundamental unit that gets reorganized during the transformation.
Problem Simplification and Explanation
Simpler Terms:
 We have a tree. Think of it like a family tree, but each person can only have up to two children.
 We want to rearrange this tree into a circle of friends where each friend knows only the next and the previous friend in the circle.
 We need to do this rearrangement without breaking the original family tree apart and using it to build the circle.
Key Concepts:
 Tree: Your initial “family tree.”
 Circle of Friends: The end goal.
 Rearrangement: The action of going from the tree to the circle.
How They Interact:
 Start with the family tree.
 Rearrange its structure so it becomes a circle of friends. Each friend should only know their next and previous friend.
 Do this while altering the original family tree, without making a new circle from scratch.
Metaphor:
Imagine you’re the director of a play. Initially, your actors stand in a hierarchical formation (the family tree) based on their roles. Now, you want to make them hold hands and form a circle for the finale. They should still maintain the order (smallest to largest based on some criteria like age). The actors are not allowed to leave the stage (inplace transformation), and they must know who stands next to them on both sides (predecessor and successor).
So, your task is to go from this hierarchical formation (tree) to a circle (doublylinked list) without changing the set (doing it inplace).
Constraints
Specific Characteristics for Efficient Solutions:
Number of Nodes: The tree has a node range of [0, 2000]. This is a limited set, allowing algorithms that are not necessarily constant time but perhaps logarithmic or linear.
Unique Values: All tree values are unique, which means we don’t have to worry about duplicate values when rearranging. This simplifies sorting.
Value Range: Node values range between 1000 and 1000. The limited range could be useful for sorting or other optimizations, though it’s less likely in this specific context.
Binary Search Tree (BST): The tree is already sorted in a specific way. Traversal in order will automatically give you sorted data, which can be used directly to create the doublylinked list.
InPlace Transformation: The requirement to do this “inplace” means you don’t need extra memory proportional to the number of tree nodes, which could be advantageous for memory optimization.
DoublyLinked List: Knowing that the end result has to be a doublylinked list gives us insights into the data structure we need to end up with. Since it’s circular, we know the start and end must be connected.
Patterns:
 Inorder traversal of the BST will yield a sorted set of nodes, which can be directly linked to form the doublylinked list.
Exploiting these characteristics can lead to more efficient algorithms and lower computational complexity.
Key Insights from Constraints:
Limited Node Count: With a maximum of 2000 nodes, solutions with linear or even logarithmic time complexity are feasible.
Unique Node Values: This eliminates the need to deal with duplicates during sorting or rearrangement, simplifying the problem.
Value Range: While the node values range from 1000 to 1000, it’s not likely to be directly exploitable for this problem. However, it does eliminate the need to consider edge cases with extremely large or small numbers.
InPlace Requirement: This constraint suggests that an efficient solution should avoid using additional memory proportional to the number of nodes. It guides us to look for algorithms that reuse existing node memory.
These constraints give us both the boundary conditions to consider and offer hints about what types of solutions may be most effective or efficient.
Case Analysis
Certainly. Here are additional examples or test cases, categorized to cover a wide range of the input space, including edge and boundary conditions:
1. Minimal Case (Edge Case)
Input: root = [] Output: [] Analysis: An empty tree should result in an empty linked list. This checks whether the solution can handle zero nodes.
2. Single Node (Edge Case)
Input: root = [1] Output: [1] Analysis: A singlenode tree should convert to a circular doubly linked list with that single node pointing to itself as both predecessor and successor.
3. Balanced BST (Common Case)
Input: root = [2, 1, 3] Output: [1, 2, 3] Analysis: A balanced tree ensures that the algorithm doesn’t favor leftheavy or rightheavy trees. The output must be a sorted circular doubly linked list.
4. Leftheavy BST (Common Case)
Input: root = [3, 2, 1] Output: [1, 2, 3] Analysis: A leftheavy tree will test if the algorithm correctly traverses left children.
5. Rightheavy BST (Common Case)
Input: root = [1, null, 2, null, 3] Output: [1, 2, 3] Analysis: Similar to the leftheavy case but for right children.
6. Random BST (Common Case)
Input: root = [10, 5, 15, 3, 7, null, 18] Output: [3, 5, 7, 10, 15, 18] Analysis: This test case doesn’t follow any particular pattern and serves as a good general test.
7. Max Nodes (Boundary Case)
Input: root = [1000, 999, 1001, …] (2000 nodes in balanced form) Output: [999, 1000, 1001, …] (sorted) Analysis: This tests the upper limit of the input size, and whether the algorithm can handle it efficiently.
8. Nodes with Negative Values (Special Case)
Input: root = [0, 1, 1] Output: [1, 0, 1] Analysis: This ensures that the algorithm can handle negative numbers, as they are within the allowed value range.
These test cases and their analyses help in understanding the edge and boundary conditions. They ensure that the solution is robust and can handle all scenarios specified within the constraints of the problem.
To visualize these cases, you can use tree diagrams for the Binary Search Trees (BST) and arrows for the resulting Circular Doubly Linked Lists (CDLL). Here’s how:
1. Minimal Case
 BST: (Empty)
 CDLL: (Empty)
2. Single Node
 BST:
1
 CDLL:
1<=>1 (circular, points to itself)
3. Balanced BST
 BST:
2 / \ 1 3
 CDLL:
1<=>2<=>3 (circular)
4. Leftheavy BST
 BST:
3 / 2 / 1
 CDLL:
1<=>2<=>3 (circular)
5. Rightheavy BST
 BST:
1 \ 2 \ 3
 CDLL:
1<=>2<=>3 (circular)
6. Random BST
 BST:
10 / \ 5 15 / \ \ 3 7 18
 CDLL:
3<=>5<=>7<=>10<=>15<=>18 (circular)
7. Max Nodes
 BST:
(Balanced tree with 2000 nodes)
 CDLL:
999<=>1000<=>1001<=>... (circular)
8. Nodes with Negative Values
 BST:
0 / \ 1 1
 CDLL:
1<=>0<=>1 (circular)
Each tree diagram represents the structure of the BST, and each arrow diagram represents the structure of the resulting CDLL.
Minimal Case: The problem allows for an empty BST, so the solution should handle this condition gracefully.
Single Node: A singlenode tree is essentially already a circular doublylinked list that points to itself. This gives us a clue that the circular doublylinked list can be thought of as an ’extension’ of the binary search tree.
Balanced BST: In a balanced tree, the linked list will have a natural sorting order without much rearrangement. This emphasizes the role of inorder traversal in preserving the order.
Leftheavy and Rightheavy BSTs: Even if the tree is not balanced, the inorder traversal ensures the correct order in the circular doublylinked list. This hints that the balance of the tree is not crucial for the problem.
Random BST: Here, the CDLL is still sorted, reinforcing the idea that as long as the BST properties are maintained, the output CDLL will be sorted.
Max Nodes: The constraint of up to 2000 nodes in the BST indicates that a solution with time complexity better than or equal to O(n) is likely necessary.
Nodes with Negative Values: The problem constraints specify that node values can be negative, zero, or positive. This tells us that our solution must handle all integers within the specified range, not just positive ones.
Key Insights:
 Inorder traversal is probably essential for a straightforward solution.
 The balance of the tree is not a significant factor.
 The solution needs to handle a variety of edge cases, including zero nodes and a single node.
 Performance constraints imply the need for an efficient algorithm, likely O(n) or better.
Identification of Applicable Theoretical Concepts
Here are some mathematical and algorithmic concepts that can simplify the problem:
InOrder Traversal: In computer science, specifically in treebased algorithms, inorder traversal is a standard method that allows us to visit tree nodes in a sorted manner. For a BST, an inorder traversal naturally produces a sorted sequence.
Circular DoublyLinked List: The operations to insert a node before or after a given node in a doublylinked list are wellunderstood and straightforward.
Pointer Manipulation: The problem asks for an “inplace” transformation, meaning we can’t use additional memory proportional to the input size. The art of changing pointers to convert the tree to a list is crucial here.
Recursion: The nature of a binary tree lends itself well to recursive approaches. Recursive algorithms can often simplify treebased problems.
Time Complexity: Given that the number of nodes is capped at 2000, an algorithm that operates in O(n) time should be acceptable. This can guide the choice of algorithm.
Graph Theory: Technically, both a tree and a linked list are specific types of graphs. Understanding the basic principles of graph theory can give a deeper understanding of the problem at hand.
Boundary Conditions: Knowing that all tree values are unique can simplify the logic since you don’t need to account for duplicate values.
These established theories and practices offer a toolbox to approach the problem efficiently and simplify its complexity.
Simple Explanation
Let’s use a metaphor. Imagine you have a family tree chart. On this chart, you have a grandparent, then their kids, and then their grandkids, all arranged in a way that makes it easy to see who is older and who is younger.
Now, instead of a family tree, you want to make a single line of family members, starting from the youngest to the oldest, and you want them to stand in a circle so that the oldest is next to the youngest.
The family tree chart is like our Binary Search Tree, where each person (or node) has links to their older and younger relatives. The single line of family members standing in a circle is like the Circular DoublyLinked List.
The challenge is to rearrange everyone from the family tree into this circle, but you have to do it without creating any new people; you can only rearrange them. That’s what the problem is asking you to do, but with computer code.
Problem Breakdown and Solution Methodology
Certainly. Let’s extend our family tree metaphor.
Steps to Solve the Problem
Initialize Variables: Create two variables,
first
andlast
, to keep track of the first and last members in the circle. They start as null.InOrder Traversal: Start from the root of the family tree and visit each node in an inorder traversal (youngest to oldest).
a. Visit Node: Every time you visit a node, disconnect it from its current position in the family tree.
b. Add to Circle: Add the visited node to the circle by updating its left and right pointers to link it to the last node in the circle. Update
first
andlast
as you go.Make Circular: After the last node is visited, connect the
first
andlast
nodes to make the list circular.Return
first
: This will be the pointer to the smallest element of the linked list.
How Steps Contribute to Solution
Initialize Variables: We need
first
andlast
to keep track of the circle’s endpoints.InOrder Traversal: This ensures we visit nodes in the sorted order.
Visit Node and Add to Circle: This step moves one person from the family tree to the circle, maintaining the sorted order.
Make Circular: This makes sure the oldest stands next to the youngest, completing the circle.
Effects of Changing Parameters
If the tree has more nodes, the inorder traversal step will take more time.
If the tree is empty (0 nodes),
first
andlast
would remain null, and you’d return null, representing an empty circle.
Example
Consider a family tree (Binary Search Tree) with these members (nodes): [4,2,5,1,3]
Initialize Variables:
first = null
,last = null
InOrder Traversal:
a. First, we visit
1
. Disconnect1
from the tree.b. Add
1
to the circle. Now,first = 1
andlast = 1
.c. Next, visit
2
, disconnect it, and add to the circle. Now,first = 1
andlast = 2
.d. Continue this for
3, 4, 5
.Make Circular: After the last node (
5
) is visited, connect5
to1
.Return
first
: Return1
as it’s the smallest element in the circle.
Now you have a sorted circle starting from 1
and going 1>2>3>4>5>1
.
Inference of ProblemSolving Approach from the Problem Statement
The key terms and how they inform the approach:
Binary Search Tree (BST): This is the data structure that stores the numbers. Because BSTs are inherently sorted, an inorder traversal will visit nodes in sorted order.
Circular DoublyLinked List: This is the data structure we aim to create. The term ‘circular’ indicates the first and last elements are connected, and ‘doublylinked’ means we can traverse in both directions.
InPlace Transformation: This indicates that no additional memory should be used. Thus, instead of creating new nodes for the linked list, we repurpose the tree nodes.
Left and Right Pointers: These are the connectors in both the tree and the list. In the tree, they refer to left and right children. In the list, they will refer to the predecessor and successor nodes.
InOrder Traversal: This term is a clue for the strategy to adopt for tree traversal. It allows us to visit tree nodes in sorted order, which is crucial for creating a sorted linked list.
Pointer to the Smallest Element: This is the output requirement, so we need to maintain a reference to the smallest element encountered during the traversal.
Constraints: The constraints (number of nodes and their value ranges) give us an idea about the problem’s scale and what kinds of optimizations might or might not be necessary.
Each of these terms or concepts provides a guideline for tackling a specific part of the problem. For example, “InOrder Traversal” directly informs us to use this method for traversing the tree to maintain sorted order. Similarly, “InPlace Transformation” restricts us from using extra memory, directing us to modify the existing nodes.
Visual aids can be extremely helpful for understanding the problem and its key properties. Here’s how you can represent each:
Binary Search Tree: Draw the tree nodes and their connections. Label each node with its value.
Circular DoublyLinked List: Draw a series of connected boxes to represent list nodes. Add two arrows for each box to indicate the ‘previous’ and ’next’ pointers. Make sure to loop the last box back to the first one to indicate the ‘circular’ nature.
InPlace Transformation: You can draw arrows from the tree nodes to their new positions in the list, indicating that the same nodes are being used.
Left and Right Pointers: In your BST diagram, label the left and right child connections. In your list diagram, label the arrows as ‘previous’ and ’next’.
InOrder Traversal: You could annotate the tree with numbers to indicate the order in which nodes will be visited. This will help visualize how the sorted list will be formed.
Pointer to the Smallest Element: In the list, mark the smallest element in some way (color it, encircle it, etc.).
Constraints: You could annotate the diagram with reminders about the number of nodes and their possible value ranges.
By creating these tables or diagrams, you’ll have a visual guide to the properties and requirements of the problem, making it easier to devise and explain your solution strategy.
How did you infer from the problem statement that this problem can be solved using ?
Simple Explanation of the Proof
I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?
Stepwise Refinement
High Level Approach:
 Step 1: Initialize pointers for tracking the previous and first nodes in the list.
 Step 2: Perform an inorder traversal of the binary search tree (BST).
 SubStep 2.1: Traverse the left subtree.
 SubStep 2.2: Update pointers to rearrange nodes as doublylinked list elements.
 SubStep 2.3: Traverse the right subtree.
 Step 3: Link the last node back to the first node to complete the circular structure.
 Step 4: Return the pointer to the smallest element in the list (i.e., the first node).
Distill Into Granular, Actionable Steps:
 Initialize Pointers: Create two pointers,
prev
andfirst
, and set them tonull
.  InOrder Traversal: Recursively visit each node.
 Go Left: Move to the leftmost node until you can’t go any further.
 Update Pointers:
 If
first
isnull
, setfirst
to the current node.  Update the
prev
node’s right pointer to point to the current node.  Update the current node’s left pointer to point to
prev
.  Set
prev
to the current node.
 If
 Go Right: Move to the right node.
 Complete the Circle: After traversal, link
prev
’s right pointer tofirst
, andfirst
’s left pointer toprev
.  Return
first
: The first node contains the smallest value and is the starting point of the circular doublylinked list.
 Initialize Pointers: Create two pointers,
Independent Parts:
 The inorder traversal of the left and right subtrees can be considered independent within the scope of each node’s operation.
Repeatable Patterns:
 The pattern of going left, updating pointers, and then going right is a repeatable pattern encountered at each node during inorder traversal.
By breaking down the problem this way, you can concentrate on smaller, more manageable tasks. These tasks often involve recurring patterns, like the inorder traversal, that you’ll perform multiple times to arrive at the solution.
Solution Approach and Analysis
Detailed Approach to Solving the Problem
Initialize Pointers:
 Create two pointers,
prev
andfirst
, both set tonull
.
 Create two pointers,
InOrder Traversal:
 Visit each node of the BST in inorder (Left, Root, Right).
2.1 Go Left:
Traverse to the leftmost node until you hit a null pointer. Think of this as walking all the way to the left edge of a garden.2.2 Update Pointers:
 If
first
isnull
, setfirst
to the current node. This is like marking the first tree you come across.  If
prev
is notnull
, link its right pointer to the current node. Also, set the current node’s left pointer toprev
. Imagine this as making a path between adjacent trees.  Update
prev
to the current node.
2.3 Go Right:
Move to the right node. This is like walking to the next tree on the right.Complete the Circle:
 After the traversal, link
prev
’s right pointer tofirst
, andfirst
’s left pointer toprev
. Think of this as making a complete loop in the garden.
 After the traversal, link
Return the First Node:
 The
first
node contains the smallest value and serves as the entry point to the circular doublylinked list.
 The
Impact of Problem Parameters on the Solution:
 Number of Nodes: The time complexity would be directly proportional to the number of nodes. A larger tree would require more time for traversal.
 Tree Structure: A balanced tree would result in better performance compared to a skewed tree.
Example Cases:
Let’s consider an example with a BST having values [4, 2, 5, 1, 3]
.
 Initialize
prev
andfirst
tonull
.  Start inorder traversal from the root (node with value
4
). Go left till you reach the node with value
1
.  Set
first
to this node (1
) and updateprev
to1
.  Move right, you find the node
2
. Link1
and2
. Updateprev
to2
.  Continue this process for nodes
3
,4
, and5
.
 Go left till you reach the node with value
 After reaching the last node (
5
), complete the circle by linking5
to1
.  Return
first
(node with value1
).
The output will be a circular doublylinked list with nodes 1, 2, 3, 4, 5
.
Identify Invariant
The invariant in this problem is the ordered nature of the nodes in both the Binary Search Tree (BST) and the resulting circular doublylinked list. In both structures, elements are arranged in ascending order.
BST Invariant: In a BST, for any given node
N
, all elements in the left subtree are smaller thanN
, and all elements in the right subtree are greater thanN
.Circular DoublyLinked List Invariant: In the circular doublylinked list, each node’s value is greater than its predecessor and smaller than its successor, maintaining the sorted order. The ‘circular’ aspect ensures that the last element connects back to the first, completing the loop.
These invariants help us maintain the sorted nature of the elements throughout the transformation process. The inorder traversal of the BST ensures that we visit nodes in ascending order, allowing us to construct the sorted circular doublylinked list in one go, while maintaining the invariants.
Identify Loop Invariant
The loop invariant in this problem would depend on the algorithm used for the transformation. However, assuming an inorder traversal, a possible loop invariant could be:
“At the beginning of each iteration of the loop, all nodes visited so far are already part of a sorted doublylinked list segment, with the ’left’ pointers pointing to their predecessors and the ‘right’ pointers pointing to their successors.”
This loop invariant ensures that the partial doublylinked list formed at any stage during the loop’s execution is sorted and maintains the predecessor and successor relationships. The loop continues to add nodes to this list, maintaining its sorted order, until the traversal is complete and the entire BST is transformed into a sorted circular doublylinked list.
Invariant and loop invariant are different for this problem.
Loop Invariant: A loop invariant is specific to a loop in your code. It’s a condition that holds before and after each iteration of the loop. In the context of this problem, the loop invariant could be related to the state of the partial doublylinked list as you go through the loop during inorder traversal, as previously described.
Invariant: An invariant in the context of the problem could be a broader condition that holds true throughout the algorithm. For example, an invariant might be that “All the nodes encountered so far are less than or equal to any nodes yet to be encountered during the inorder traversal.”
Invariants provide a general level of assurance across the entire algorithm or system, while loop invariants are specifically tied to individual loops within that system.
Thought Process
Basic Thought Process and Steps
Recognize the Domain: This is a problem about Binary Search Trees and DoublyLinked Lists, which are data structures in computer science.
Identify Components:
 Binary Search Tree (BST)
 Circular DoublyLinked List
 Inplace transformation
 Understand Constraints and Requirements:
 The transformation should be inplace.
 Number of nodes is in the range [0, 2000].
 All node values are unique.
 Direction from the Problem Statement:
 Cues: “in place”, “predecessor and successor”, “return the pointer to the smallest element.”
 These cues suggest an inorder traversal of the BST to maintain the sorted order, and manipulating the pointers to convert it into a doublylinked list.
 Insights:
 Since we are dealing with inplace operations, we can use the left and right pointers of the tree nodes as the previous and next pointers of the doublylinked list nodes.
 The circular condition implies the last node should point back to the first node.
Code
PENDING TEST Here’s the Python3 code to implement the solution.


This code maintains the sorted nature of BST while converting it into a Circular DoublyLinked List. The time complexity is (O(N)), where (N) is the number of nodes.
Establishing Preconditions and Postconditions
Parameters
 Inputs: The input to the method is
root
.  Types:
root
is of typeNode
, whereNode
is a custom class representing a node in the BST.  Representation: In the context of the problem,
root
represents the root node of the Binary Search Tree.
Preconditions
 State: Before calling this method, the
root
should represent a valid BST, or beNone
for an empty tree.  Constraints:
 Number of nodes is between [0, 2000].
 Each node value is between 1000 and 1000.
 All node values are unique.
 Program State: No specific program state is required.
Method Functionality
 Expected to Do: The method is expected to convert the BST into a sorted Circular DoublyLinked List inplace.
 Interaction: It traverses the BST in an inorder fashion and rearranges the pointers to make it a Circular DoublyLinked List.
Postconditions
 State: After the method returns, the BST is transformed into a Circular DoublyLinked List.
 Return Value: The return value is a pointer to the smallest element of the newly formed list.
 Side Effects: The original BST is modified.
Error Handling
 Preconditions Not Met: The method handles empty trees by returning
None
.  Response: It doesn’t throw an exception or return a special value. If the input BST is empty, it returns
None
.
Problem Decomposition
Problem Understanding
In Own Words: The problem is to convert a Binary Search Tree (BST) into a Circular DoublyLinked List while keeping it sorted. This has to be done inplace, meaning we modify the existing tree rather than creating a new structure. We should return a pointer to the smallest element.
Key Components and Requirements:
 Convert BST to DoublyLinked List
 Keep it sorted
 Do it inplace
 Return pointer to the smallest element
Initial Breakdown
 Major Parts:
 Traversal of BST
 Link node conversion
 Making the list circular
Subproblem Refinement
 Traversal of BST:
 Traverse left subtree
 Visit root
 Traverse right subtree
 Link Node Conversion:
 Change left and right pointers to act as previous and next pointers in a doublylinked list.
 Making the List Circular:
 Connect the last and first elements
Task Identification
 Traversal:
 Repeated for each node and subtree
 Node Conversion:
 Done for each node visited
 Making Circular:
 Done once, after all nodes are visited
Task Abstraction
 Traversal: General enough to be used for any BST
 Node Conversion: Specific to this problem, yet reusable in a similar context
 Making Circular: Specific to making doublylinked lists circular, but reusable in that context
Method Naming
 Traversal:
inOrderTraversal
 Node Conversion:
convertToDLL
 Making Circular:
makeCircular
Subproblem Interactions
 Order and Dependencies:
 Traversal must be done first, as it sets the stage for node conversion.
 Node conversion is done during traversal, making these steps intertwined.
 Making the list circular is the last step after the entire list is formed.
By breaking down the problem into these subproblems and tasks, we can proceed with a systematic approach to solving it.
From Brute Force to Optimal Solution
Brute Force Solution
Approach
 Perform inorder traversal of the BST, and store the elements in an array.
 Create a new Circular Doubly Linked List using the sorted array.
 Make the list circular by connecting the last and first elements.
Code for Brute Force
Here’s a simplified Python code for the bruteforce solution.


Inefficiencies
Time Complexity:
 Traversing the BST: O(N)
 Building a new list: O(N)
 Total: O(N)
Space Complexity:
 Storing elements in an array: O(N)
 Creating a new list: O(N)
 Total: O(N)
Not InPlace: We create a new structure rather than modifying the existing one.
Optimized Solution
Approach
 Perform inorder traversal of the BST.
 Convert each visited node to a doublylinked list node during traversal.
 Make the list circular.
Optimization Steps
 InPlace Conversion: Instead of using an array and creating a new list, we modify the existing BST nodes to become DLL nodes during the traversal.
Code for Optimized Solution


Time and Space Complexity
Time Complexity: O(N), same as bruteforce but with a single pass.
Space Complexity: O(1), we do everything inplace.
By converting the tree inplace during the inorder traversal, we save on space and also adhere to the problem constraint of performing the conversion inplace.
Code Explanation and Design Decisions
1. Initial Parameters
root
: It’s the root node of the Binary Search Tree (BST). This is where our traversal starts.prev
: It’s the previous node visited during the inorder traversal. This is essential to link the current node with the previous one in the Doubly Linked List (DLL).
2. Primary Loop or Iteration
The primary iteration is the recursive flattenDLL
function calls. Each iteration traverses a node in the BST and modifies it to be part of the DLL.
 Iteration over
root.left
: Traverses the left subtree, linking all its elements into the DLL.  Iteration over
root
: Modifies the root node to fit into the DLL, linking it with the previous node.  Iteration over
root.right
: Traverses the right subtree, continuing the linking.
3. Conditions or Branches
if not root: return prev
: Checks if the node is null. This signifies we’ve reached a leaf node and should go back.if prev: prev.right = root
: Checks if a previous node exists. This condition ensures that we can link the current node with the previous one without encountering a null reference error.
4. Updates or Modifications
root.left = prev
: Links the current node’s left pointer to the previous node.prev = root
: Updates theprev
node to be the current root. This is critical for the next iteration, as this current node will serve as the previous node for its successor in the DLL.
5. Invariant
The invariant is the inorder sequence of nodes in the BST. Throughout the code, we maintain this order in the DLL, satisfying the problem’s constraints. The prev
node always points to the last node visited in this inorder traversal, ensuring that every node links correctly in the DLL.
6. Final Output Significance
The final output is the head of the DLL, which is the smallest element in the BST. This output meets the problem’s requirements of converting the BST into a sorted, circular DLL. The circular link is established, and the ordering is maintained, thereby satisfying all constraints and objectives.
Coding Constructs
1. HighLevel ProblemSolving Strategies
The code uses recursion and inorder traversal to solve the problem. It also employs pointer manipulation to convert the binary tree into a doublylinked list.
2. Explaining to a NonProgrammer
Imagine you have a family tree chart. This code rearranges that family tree into a single line of people, sorted by age, while still keeping track of who comes before and after each person. Plus, the line loops back on itself, so the oldest and youngest are also connected.
3. Logical Elements
 Conditional checks (
if
statements): Determine whether a certain condition is met and decide the next course of action.  Loops (via recursion): Go through each item in the data structure.
 Pointers: Reference specific items in the data structure for modification.
4. Algorithmic Approach in Plain English
Start at the top of the family tree (root). Move to the leftmost person (youngest generation), then go up one generation at a time, linking each person to the one just older than them. Continue doing this until you reach the oldest person. Finally, link the oldest and youngest to complete the circle.
5. Key Steps or Operations
 Navigate to the leftmost (youngest) person: To ensure the list starts with the smallest value.
 Link the current person to the previous person: To maintain the sorted order.
 Move to the right (older generation): To continue the inorder traversal and linking.
6. Algorithmic Patterns
 Tree Traversal: The code uses inorder traversal to visit each node in the tree.
 Recursion: The code calls itself to solve smaller instances of the same problem.
 Pointer Manipulation: The code rearranges pointers to convert the tree structure into a linked list.
Language Agnostic Coding Drills
1. Distinct Coding Concepts
Variable Declaration and Initialization: Understanding how to declare and initialize variables.
Conditional Statements: Employing
if
statements to make decisions in code.Looping Mechanisms: Using loops to iterate through a collection or repeat tasks.
Functions and Methods: Creating reusable blocks of code.
Recursion: Function calling itself to solve a problem in parts.
Data Structures: Understanding and implementing structures like trees and linked lists.
Pointer Manipulation: Modifying links between nodes to rearrange or transform data structures.
Inorder Tree Traversal: Walking through a binary tree in a specific order.
Algorithm Optimization: Refining algorithms for better performance.
2. Difficulty Level and Descriptions
Variable Declaration and Initialization: Easiest. Basics of storing data.
Conditional Statements: Easy. Adds logic to the program.
Looping Mechanisms: Moderate. Adds complexity through repetition.
Functions and Methods: Moderate. Requires understanding of scope and modularity.
Data Structures: More Challenging. Needs conceptual understanding of how data is stored and retrieved.
Recursion: Challenging. Requires a good grasp of stack frames and termination conditions.
Inorder Tree Traversal: Quite Challenging. Needs understanding of trees and recursion.
Pointer Manipulation: Difficult. Requires a deep understanding of memory and data structures.
Algorithm Optimization: Most Difficult. Requires mastery of prior concepts and analytical skills.
3. ProblemSolving Approach
Variable Declaration and Initialization: Start by defining the necessary variables like pointers to keep track of the current and previous nodes.
Conditional Statements: Use conditionals to check if you’ve reached a leaf node or the end of a branch.
Looping Mechanisms: Initially, think about how a loop could iterate through the tree. This will later be replaced by recursion.
Functions and Methods: Encapsulate the logic in a function that you can call repeatedly.
Data Structures: Understand the tree and linked list structures you’ll be working with, and how they are defined.
Recursion: Replace the loop with a recursive function that traverses the tree to perform the task on each node.
Inorder Tree Traversal: Implement inorder traversal within the recursive function to ensure you visit each node in the desired order.
Pointer Manipulation: As you traverse, modify the pointers to change the tree into a linked list.
Algorithm Optimization: Lastly, refine the solution for efficiency, making sure the conversion happens inplace and optimally.
Targeted Drills in Python
1. Python Coding Drills for General Concepts
Variable Declaration and Initialization


Conditional Statements


Looping Mechanisms


Functions and Methods


Recursion


Data Structures


2. ProblemSpecific Drills
Inorder Tree Traversal


This drill is essential because you need to know how to traverse a tree in a specific order to solve our initial problem.
Pointer Manipulation


Pointer manipulation is key to efficiently altering data structures, crucial in transforming a tree to a linked list, for instance.
3. Integrating the Drills to Solve the Problem
Variable Declaration and Initialization: Start by defining necessary variables. Initialize tree nodes and pointers for traversal.
Data Structures: Build the tree and linked list structures you’ll be working with.
Functions and Methods: Write functions that will encapsulate smaller tasks like traversing a tree or converting a subtree to a linked list.
Conditional Statements: Inside these functions, implement conditionals to guide the traversal or transformation process.
Recursion: Implement recursive tree traversal within your functions. Use the inorder traversal method.
Looping Mechanisms: If you decide to implement an iterative version, use loops for traversal.
Pointer Manipulation: Within your recursive function, perform necessary pointer manipulations to modify the tree into a linked list.
By following this sequence, you should be able to integrate these drills into a complete solution for the initial problem. Each drill corresponds to a portion of the task and contributes to the overall solution.
Q&A
Similar Problems
Certainly, based on the coding concepts and strategies that were relevant to the problem you’re working on, here are 10 problems from LeetCode that employ similar problemsolving techniques:
Two Sum
 Similarity: Uses hash tables for quick lookup, similar to how you might use a dictionary to store traversed nodes in a tree.
Reverse Linked List
 Similarity: Requires pointer manipulation, much like you would need when transforming a binary tree to a doubly linked list.
Maximum Depth of Binary Tree
 Similarity: Requires recursive traversal of a tree, which is a critical skill you needed for your original problem.
Valid Parentheses
 Similarity: Requires a stack to keep track of elements in a specific order, analogous to how you may use a stack in tree traversal.
Merge Two Sorted Lists
 Similarity: Requires understanding of pointers and how to manipulate linked list nodes, skills that are transferable to manipulating tree nodes.
Invert Binary Tree
 Similarity: Involves recursive traversal and pointer manipulation of tree nodes, which closely aligns with your original problem’s requirements.
Remove Duplicates from Sorted Array
 Similarity: Requires efficient inplace manipulation of a data structure, a technique that could also be used in tree transformations.
Binary Tree Level Order Traversal
 Similarity: Requires traversal of a binary tree, possibly using a queue, which could be a strategy for your original problem too.
Climbing Stairs
 Similarity: Involves solving a problem through recursion or dynamic programming, resembling how you might handle different subtrees recursively.
Palindrome Linked List
 Similarity: Requires twopointer technique to traverse a linked list, a concept that can be adapted for tree to linkedlist transformations.
These problems were chosen because they incorporate essential coding concepts like recursion, pointer manipulation, data structure manipulation, and traversal strategies, which are also key to solving your original problem.