Find Mode in Binary Search Tree
First, to find the modes in a binary search tree (BST), we need to traverse the tree, count the occurrences of each value, and then identify the maximum frequency. Due to the properties of the BST, an inorder traversal will yield the values in nondecreasing order, making it easier to track frequencies.
Python solution:


Here, the function inorderTraversal
goes through the tree in an inorder fashion. For each visited node, it increases the corresponding count in the self.count
dictionary and updates self.maxCount
if necessary.
After the traversal, the function findMode
iterates over the self.count
dictionary. For each item with a count equal to self.maxCount
, it appends the key to the self.modes
list, which is then returned as the result.
Identifying Problem Isomorphism
“Find Mode in Binary Search Tree” requires identifying the most frequently occurring element(s) in a Binary Search Tree (BST). It uses inorder traversal of the BST and keeps track of the count of the current number and the maximum count seen so far, in order to identify the mode(s).
An isomorphic problem to this is “Vertical Order Traversal of a Binary Tree” (LeetCode #987). In this problem, you’re given a binary tree (not necessarily a BST), and you’re asked to return the vertical order traversal of its nodes’ values.
In both problems, you need to traverse the tree and aggregate node values based on some criteria: for “Find Mode in Binary Search Tree”, you aggregate counts for each value to find the mode(s); for “Vertical Order Traversal of a Binary Tree”, you aggregate values by their vertical positions.
It is an approximate isomorphism, because while the “Find Mode in Binary Search Tree” uses the property of a BST to make the traversal simpler, the “Vertical Order Traversal of a Binary Tree” requires a more complex traversal method that can track both vertical and horizontal position. Yet, both problems involve traversing a tree and aggregating values in some way, hence can be considered as a form of isomorphism.


Problem Classification
This problem falls into the domain of Computer Science and more specifically, the subdomain of Data Structures and Algorithms.
What:
A binary search tree (BST) with possible duplicate values.
The need to identify and return the mode(s) of the BST, which are the values that appear most frequently in the tree.
The problem has specific constraints related to the properties of a BST (keys in the left subtree are less than or equal to the root’s key, keys in the right subtree are greater than or equal to the root’s key).
This problem is an instance of the Traversal and Analysis of Trees problem type. This is because the solution involves traversing the tree, likely using an algorithm like depthfirst or breadthfirst search, and analyzing the values to identify the mode(s). The problem also involves understanding and utilizing the properties of BSTs to optimize the traversal and analysis. The problem is focused on the manipulation and understanding of the BST data structure and hence is a Data Structure problem.
This problem might involve concepts such as tree traversal algorithms, frequency counting, and possibly recursion if a recursive tree traversal method is used. The challenge and complexity of the problem come from efficiently traversing the BST and keeping track of value frequencies in a way that allows easy identification of the mode(s).
Language Agnostic Coding Drills
Concepts found in the code:
a. Basic Data Types and Data Structures  This includes the use of lists and dictionaries in Python. This concept is fundamental in understanding the use of variables, storage, and retrieval of data.
b. Conditional Statements  The
if
statement is used to make decisions based on certain conditions. In this case, it’s used to check if a node value is already in the dictionary.c. Function Definition  A nested function
find
is defined and called within the main function. This is a part of the broader concept of organizing code into reusable blocks.d. Recursion  The
find
function calls itself to traverse the binary tree. This is a part of recursive function calling which is a significant concept in computer science.e. Iterating over a Dictionary  The
for
loop is used to iterate over keyvalue pairs in a dictionary.List of coding concepts in order of increasing difficulty:
a. Basic Data Types and Data Structures  Beginner level, essential for almost any coding task.
b. Conditional Statements  Beginner level, used for controlling the flow of the program.
c. Function Definition  Intermediate level, important for code organization and reusability.
d. Iterating over a Dictionary  Intermediate level, a common pattern in many programming tasks that involve data manipulation.
e. Recursion  Advanced level, as it requires understanding how function calls are stored on the call stack and how base and recursive cases work together.
Problemsolving approach:
The approach to solving the problem follows a few key steps:
a. Tree Traversal  Using a depthfirst search (DFS) method, we traverse through each node in the binary tree. This is achieved by recursively calling the
find
function on the left and right children of each node.b. Frequency Counting  As we visit each node, we update a dictionary (
dic
) with the node’s value as the key and the frequency as the value. This provides us a way to keep track of the occurrence of each value in the binary tree.c. Finding the Maximum Frequency  We find the maximum frequency from our dictionary. This represents the frequency of the mode(s) in our BST.
d. Identifying the Modes  We iterate over our dictionary to find all values that have a frequency equal to the maximum frequency identified in the previous step. These values are our modes, and we append them to our result list (
lst
).Each of these steps involves a concept identified above. Starting from simple data structures and conditional statements, we gradually introduce more advanced concepts like recursion and dictionary iteration to solve the problem. It’s an example of how different programming concepts can work together to solve complex problems.
Targeted Drills in Python
Python drills for each concept:
a. Basic Data Types and Data Structures:
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# Lists my_list = [1, 2, 3, 4, 5] print(my_list) # Dictionaries my_dict = {'one': 1, 'two': 2, 'three': 3} print(my_dict)
b. Conditional Statements:
1 2 3 4 5
x = 10 if x > 5: print("x is greater than 5") else: print("x is less than or equal to 5")
c. Function Definition:
1 2 3 4
def greet(name): print(f"Hello, {name}!") greet("Python Learner")
d. Iterating over a Dictionary:
1 2 3
my_dict = {'one': 1, 'two': 2, 'three': 3} for key, value in my_dict.items(): print(f"Key: {key}, Value: {value}")
e. Recursion:
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def factorial(n): if n == 1: return 1 else: return n * factorial(n1) print(factorial(5)) # Output: 120
Problemspecific drills:
a. Tree Traversal  Recursive function to traverse a binary tree:
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def traverse(node): if node is not None: print(node.val) traverse(node.left) traverse(node.right)
b. Frequency Counting  Counting the frequency of items in a list:
1 2 3 4 5 6 7 8
def count_frequency(my_list): freq_dict = {} for item in my_list: if item in freq_dict: freq_dict[item] += 1 else: freq_dict[item] = 1 return freq_dict
These drills are essential for our problem because we need to traverse a binary tree, track the frequency of elements (node values), and identify elements with the highest frequency.
Integration of drills for the final solution:
a. First, we use the tree traversal drill to visit each node in the BST.
b. As we traverse, we apply the frequency counting drill to track the occurrence of each node value. We store this information in a dictionary.
c. Once we’ve traversed the entire tree, we then determine the maximum frequency from our dictionary. This is done by finding the maximum value in the dictionary.
d. Finally, we iterate over the dictionary (using the dictionary iteration drill), and for each keyvalue pair, we check if the value matches the maximum frequency. If it does, we append the key to our result list.
The result list, which contains all the modes of the BST, is then returned as the final solution. The ordered combination of these drills successfully solves the problem.