Check If Two Expression Trees are Equivalent

The key to this problem is understanding that the order of operands doesn’t matter in an expression only involving addition operations (due to the commutative property of addition). Therefore, two binary expression trees are equivalent if they contain the same count of each variable, regardless of structure.

To solve this problem, you could perform a depth-first search (DFS) on each tree and count the occurrences of each variable in a dictionary or similar data structure, then compare the two dictionaries to see if they are the same.

Python solution:

from collections import Counter

class Solution:
    def checkEquivalence(self, root1: 'Node', root2: 'Node') -> bool:
        def dfs(node: 'Node', counter: Counter):
            if node is None:
            if node.val != '+':
                counter[node.val] += 1
            dfs(node.left, counter)
            dfs(node.right, counter)

        counter1, counter2 = Counter(), Counter()
        dfs(root1, counter1)
        dfs(root2, counter2)
        return counter1 == counter2

In this code, we use Counter from collections to store the counts of variables, and a helper function dfs to perform the depth-first search on each tree. The dfs function updates the counter for each non-’+’ value it encounters in the tree. Finally, we compare the two counters to see if they are the same. If they are, we return True, indicating that the two trees are equivalent. If not, we return False.

10 Prerequisite LeetCode Problems

This problem is about binary tree traversals, expression evaluation, and equivalence of expressions. Here are 10 problems that involve similar concepts:

  1. 94. Binary Tree Inorder Traversal: You need to return the inorder traversal of a binary tree.

  2. 144. Binary Tree Preorder Traversal: This problem requires you to return the preorder traversal of a binary tree.

  3. 145. Binary Tree Postorder Traversal: This problem involves the postorder traversal of a binary tree.

  4. 150. Evaluate Reverse Polish Notation: You need to evaluate an arithmetic expression given in Reverse Polish Notation, which involves concepts of expression evaluation.

  5. 652. Find Duplicate Subtrees: You are required to find duplicate subtrees in a binary tree, which involves tree comparison.

  6. 872. Leaf-Similar Trees: This problem requires you to compare the leaf sequences of two binary trees.

  7. 100. Same Tree: This problem asks whether two binary trees are identical.

  8. 951. Flip Equivalent Binary Trees: You need to determine whether two binary trees are flip equivalent, i.e., whether you can make them equal by flipping some nodes.

  9. 993. Cousins in Binary Tree: This problem is about identifying sibling nodes in a binary tree.

  10. 958. Check Completeness of a Binary Tree: You are required to check whether a binary tree is complete.

As for the follow-up question: If the tree also supports the ‘-’ operator, the problem would become more complex. In this case, you cannot simply compare the counts of variables because subtraction is not commutative. You might need to evaluate the expressions in the trees with different variable values and compare the results to decide whether the two trees are equivalent.

Clarification Questions

What are the clarification questions we can ask about this problem?

Problem Analysis and Key Insights

What are the key insights from analyzing the problem statement?

Problem Boundary

What is the scope of this problem?

How to establish the boundary of this problem?

Problem Classification

This problem involves evaluating and comparing binary expression trees, so it falls under the domain of trees and mathematical expressions.

The key ‘What’ components are:

  • Binary expression trees with node values of operators and operands
  • Trees follow structure rules (internal nodes have 2 children)
  • Evaluating trees based on operator and operand values
  • Comparing evaluation results between 2 trees
  • Determining if trees are equivalent by evaluating to same value

Based on these components, we can further classify this problem as:

  • Tree theory - Modeling structure and relationships between node parents/children
  • Expression evaluation - Evaluating mathematical expressions represented as trees
  • Value comparison - Comparing results of evaluating 2 expression trees
  • Equivalence checking - Determining if 2 trees are equivalent by evaluating to equal values

So in summary, this is an equivalence checking problem involving evaluating and comparing binary expression trees. It relies heavily on concepts from tree data structures and mathematical expressions. The core challenge is traversing and evaluating the 2 trees correctly to check if they result in the same value.

Distilling the Problem to Its Core Elements

This problem is based on the fundamental concepts of tree traversal and expression evaluation.

At its core, it involves traversing two tree structures representing mathematical expressions, evaluating each expression, and checking if the results are equal.

In simple terms, I would describe it as:

“Given two math expressions built using additions and variables, evaluate each one fully and check if they result in the same number.”

The core problem is determining if two expressions are functionally equivalent, not the actual expression values. We can simplify it to:

“Do these two expression trees evaluate to the same result?”

The key components are:

  • Modeling the trees
  • Traversing the trees
  • Evaluating the expressions
  • Comparing the results

The minimal set of operations is:

  • Build tree structures
  • Recursively traverse and evaluate each tree
  • Compare final evaluation values
  • Return true if equal, false otherwise

So in essence, this problem focuses on using tree structures to model expressions, evaluating them correctly through traversal, and comparing the results - rather than complex expression calculations.

Visual Model of the Problem

Here are some ways we could visualize the problem statement:

  • Show two sample binary expression trees using standard tree structure and operand/operator labels. Highlight equivalent subtrees.

  • Step through an example evaluation of each tree, writing out the expression as you traverse each node. Show how they evaluate to the same value.

  • Animate traversing both trees in parallel, showing the evaluation process and resulting values at each step. Illustrate how the final values are equal.

  • Use a flow chart to show the high-level process - build trees, traverse/evaluate, compare values, return true/false result.

  • Create small example expression trees that are equivalent and non-equivalent. Show evaluations that lead to same and different results.

  • Visualize invalid examples like trees with different structures or multiple operators.

  • Use visual encoding like color coding nodes and edges based on contribution to the final value.

  • Leverage tree visualization libraries to interactively demonstrate evaluations and comparisons.

The goal is to leverage multiple visual methods like examples, flow charts and animations to demonstrate the concepts of expression tree evaluation and equivalence checking in an intuitive way. Both valid and invalid cases help illuminate the problem.

Problem Restatement

Here’s how I would paraphrase the problem statement in my own words:

We are given two binary trees that represent mathematical expressions containing variables and the addition operator. Each internal tree node represents an addition operation and the leaf nodes are operands (variables).

Our goal is to evaluate each expression tree fully, by traversing the nodes and calculating the result based on the operator and operand values. Once we have the evaluated result for each tree, we need to compare them.

If the two expression trees evaluate to the same numerical value, we should return true, indicating the trees are equivalent expressions. If the evaluation results are different, we return false as the trees are not equivalent.

Some key points:

  • The input trees follow the standard rules for binary expression trees

  • We need to recursively traverse and evaluate both trees fully

  • The specific variable values don’t matter, we just need the final numerical result

  • We are only comparing the final numeric evaluation results, not the tree structures

  • Only the ‘+’ operator is used, no other mathematical operations

So in summary, we must traverse the two input expression trees, evaluate them into numeric results, compare the results, and return true if they match or false if they differ. The core is evaluating both trees correctly by traversing all nodes.

Does this help summarize the problem’s essence and constraints? Let me know if any part needs more clarification or I missed anything important.

Abstract Representation of the Problem

Here is one way to formulate an abstract representation of this problem:

Let T1 and T2 be two rooted tree data structures with nodes N1 and N2 respectively.

Each node N has:

  • A value V (operator or operand)
  • Left child CL and right child CR subtrees

We define an evaluation function eval(T) that takes a tree T and returns a numeric value V:

  • If N is a leaf node, V is the operand value
  • If N is internal, V is the result of applying N’s operator to CL and CR subtree evals

The problem is then:


  • Rooted trees T1 and T2


  • If eval(T1) = eval(T2)


  • True if eval(T1) = eval(T2)
  • False otherwise

This frames the problem in a generic way using abstract tree structures and an evaluation function that operates on them. The specifics of a binary expression tree and addition operator are encapsulated.

The key aspects are the recursive evaluation function that computes a value for each tree, and comparing these values for equality.

Does this help provide a language and domain agnostic representation? Let me know if you would modify or clarify the abstraction.


Here are some key terms and concepts for understanding this problem:

Binary expression tree - A tree where each internal node represents an operator and leaf nodes are operands. Used to model mathematical expressions.

Tree traversal - Visiting all nodes in a tree in a systematic order, e.g. depth-first or breadth-first. Required to evaluate the expressions.

Recursion - A technique where a function calls itself to repeat the same logic. Useful for traversing trees recursively.

Operator node - An internal tree node holding an operator like ‘+’ to apply to its operands.

Operand node - A leaf node in the tree holding a variable value or number.

Evaluation - Computing a final value for the expression tree based on its operators and operands.

Equivalence - Two expression trees have the same functionality if they evaluate to equal values, even if structured differently.

Subtree - A smaller tree nested as part of the overall tree structure. Subtrees are independently evaluated.

Some other relevant concepts are binary trees, abstraction, mathematical expressions, comparison, depth-first vs breadth-first traversal, tree isomorphism, etc.

But the key terms like binary expression tree, traversal, recursion, operator/operand nodes, evaluation and equivalence help understand and solve this problem.

Problem Simplification and Explanation

Here’s one way I can break down and explain this problem in simpler terms:

Key Concepts:

  • Tree - The inputs are tree data structures with branches and leaf nodes

  • Expression - The trees represent mathematical expressions with operators and operands

  • Traversal - We need to traverse all nodes in the trees systematically

  • Evaluation - Traversing lets us calculate the expression’s value

  • Equivalence - We compare results to determine if expressions are equal


Think of the trees as complicated math formulas. The internal nodes are like +, -, x, / operators while the leaf nodes are numbers or variables.

We want to fully simplify both formulas step-by-step, following the order of operations, until we get a single number result for each one.

Once we have both results, we just need to check if they are exactly equal. If so, the formulas are equivalent even if structured differently.

It’s like simplifying ((2 + 3) x 5) and (5 x (2 + 3)) - they give the same result of 25 even though structured differently. We just evaluate both and compare numbers.

In summary, the key is modeling the tree structure, recursively simplifying it through traversal, evaluating to a single value, and comparing values. Breaking it into these distinct steps helps understand each aspect.


Here are some characteristics of the problem that could help optimize or simplify the solution:

  • Tree size is limited to 4999 nodes. Small enough for brute force traversal.

  • Node values are limited (either ‘+’ or variables). Can encode these for fast checking.

  • No nested expressions or complex operators. Simpler evaluation logic.

  • ‘+’ is commutative. Order of evaluation does not matter.

  • Identical tree shape. Don’t need complex isomorphism checks.

  • Leaf nodes have fixed operand values. These can be precomputed.

  • No output requirements like printing the expression. Only care about equality.

  • Intermediate evaluation results don’t matter, only the final value.

  • Can traverse in any order, as long as all nodes are covered.

  • Subtrees can be evaluated independently and results cached.

The limited tree size means we likely don’t need to optimize data structures or logic much. Simple brute force traversal would suffice.

We also don’t need to deal with order of operations or optimizations in evaluation order due to the single operator. Caching/pruning optimizations could help speed up traversal.

Here are some key insights gained by analyzing the constraints:

  • Small tree size allows brute force approaches without needing complex optimizations.

  • Fixed operator and operands simplifies evaluation logic compared to complex expressions.

  • Commutative property of addition means we don’t need to optimize evaluation order.

  • Identical tree shapes ensure we don’t need expensive structure comparisons.

  • Precomputing leaf nodes saves redundant operations during traversal.

  • Only caring about equality of final values, not intermediate steps, simplifies evaluation requirements.

  • Flexible traversal order gives us freedom to optimize search based on structure.

  • Independent subtrees enable caching/memoization to reuse work and prevent recomputation.

  • No output formatting needs means we can focus just on calculation and comparison.

Overall, these constraints allow us to simplify the problem space and avoid over-engineering the solution. We can use brute force approaches relying on the limited size and simple expression structure. The focus narrows to just efficient traversal and comparison rather than complex evaluation logic. The insights help prune out unnecessary problem dimensions.

Case Analysis

Here are some additional test cases covering different aspects:

  1. Identical trees

Input: Tree 1: [+, a, b] Tree 2: [+, a, b]

Output: True

Analysis: Basic case, same structure and values.

  1. Different leaf values

Tree 1: [+, a, b] Tree 2: [+, a, c]

Output: False

Analysis: Different operand values lead to different evaluation.

  1. Different structure

Input: Tree 1: [+, [+, a, b], c]
Tree 2: [+, a, [+, b, c]]

Output: True

Analysis: Order doesn’t matter since + is commutative. Checks structure differences.

  1. Overflow

Input: Tree 1: [+, 9999, 1] Tree 2: [+, 9998, 2]

Output: False

Analysis: Large values could cause overflow.

  1. Unbalanced trees

Input: Tree 1: [+,[+,a,b],c] Tree 2: [+,a,[+,b,c]]

Output: True

Analysis: Handles unbalanced/skewed trees.

Edge cases:

  • Identical trees - same structure and values
  • Overflow/underflow
  • Unbalanced tree shapes
  • Very large trees close to max size

Other cases help test commutativity, varying structures and values, and large inputs.

Here are some ideas for visualizing these test cases:

  • Use standard tree diagrams to show structure and node values

  • Mark equivalent subtrees and nodes with same color

  • Annotate leaf nodes with evaluated values

  • Show traversal order and evaluation steps on each node

  • Illustrate overflow/underflow, e.g. node values too large

  • Vary node sizes or shapes to indicate unbalanced trees

  • Keep diagrams small and simple for basic cases

  • Use multi-color edges to highlight non-equivalent parts

  • Split screen comparing two test trees side by side

  • Leverage tree visualization libraries for interactive examples

  • Animate traversals and evaluation over time for complex cases

  • Visualize invalid scenarios like mismatched trees

  • Use flow charts to show high-level logic and resulting equivalence

The goal is to leverage visual tools like diagrams, animations and colors to illustrate the key concepts of expression tree structure, evaluation and equivalence. Both valid and invalid examples help highlight subtle details.

Here are some key insights gained from analyzing these test cases:

  • Identical trees validate basic traversal and comparison works. Important baseline.

  • Varying operand values shows equivalence depends on evaluation results, not just structure.

  • Different structures highlight need to fully evaluate rather than compare shapes.

  • Overflow cases reveal importance of handling large values correctly.

  • Unbalanced trees emphasize requirement to process any valid tree shape.

  • Need to verify commutative property holds by testing different structures.

  • Large trees stress test optimization and scalability.

  • Simple examples are crucial for isolating logic and debugging.

  • Negative examples are useful for debugging subtle failures.

  • Visualization makes tree structure, traversal and evaluation more intuitive.

  • Fundamentally, the insights show equivalence relies on full evaluation and value comparison rather than structural or shape based checks. The test cases validate this core idea from multiple angles.

Identification of Applicable Theoretical Concepts

Here are some mathematical and algorithmic concepts that could help solve this problem:

  • Tree traversal algorithms like depth-first or breadth-first search allow systematically visiting all nodes

  • Recursion is a natural fit for evaluating subtrees nested within a larger tree structure

  • Mathematical expression rules guide the order of operations and operator precedence during evaluation

  • Commutativity of addition allows evaluating subtrees in any order

  • Associativity of addition ensures parenthesization doesn’t affect result

  • Caching/memoization can store intermediate subtree evaluations to avoid recomputing

  • Hash functions can map subtree states to cached values for faster lookup

  • Isomorphism algorithms can compare tree structure, but not needed here

  • Heuristics like evaluating simpler subtrees first could optimize traversal

Concepts from discrete math like graph theory, recursion, expression evaluation order, and memoization give useful techniques for traversing the tree, calculating values incrementally, and optimizing repeated computations. The problem also lends itself well to recursive solutions.

Simple Explanation

Here’s how I would explain this problem in simple non-technical terms:

Imagine you have two math equations made up of numbers, variables, and operations like add, subtract, multiply. For example:

(A + B) * C

Now imagine we don’t know what numbers A, B and C are. But we want to see if the two equations give the same result for any values of A, B and C.

I would evaluate each equation step-by-step, following the normal math rules of what to calculate first. Once I simplify it down to a single number, I can compare the two results.

If both equations give the same number at the end, then they are equivalent! Even if the equations looked a bit different originally before we simplified them.

It’s like having: (2 + 3) * 5 and 5 * (2 + 3)

These give 25 in the end after doing the math operations. So we’d say these complex equations are equal or equivalent, even though they look different before simplifying.

We just need to carefully simplify, always following the same math rules, then compare the numbers we end up with - ignoring the original structure.

Problem Breakdown and Solution Methodology

Here is a step-by-step explanation of how I would approach solving this problem:

  1. Model the input as tree data structures
  • Each node contains either an operator or operand value
  • Build out the tree structure based on node relationships
  1. Implement a tree traversal function
  • Use depth-first search (DFS) to visit all nodes systematically
  • Recursively call traversal on child nodes
  1. Evaluate nodes during traversal
  • If node is operator, compute result from children recursively
  • If node is operand, simply return value
  • Store result in node as we traverse
  1. Compare final results
  • After full traversal, compare root node values
  • Return true if equal, false otherwise

This incrementally evaluates the expressions by traversing the trees, computing and storing node values along the way. We only need the final root node values to compare.

If nodes could have complex expressions, we’d have to be careful of order of operations. With just addition, we can evaluate subtrees in any order.


       *                     *
    /     \               /     \  
  +        C            A        +
 / \                          /  \
A   B                        B    C
  1. Build tree structures

  2. Traverse left tree DFS, compute A + B = X, store in * node

  3. Traverse right tree DFS, compute B + C = Y, store in * node

  4. Compare final values X and Y, return true if equal

This shows incrementally evaluating the trees and comparing the end result. We could visualize this using step-by-step tree diagrams.

Inference of Problem-Solving Approach from the Problem Statement

Here are some key terms and how they guide my approach:

  • Tree - Tells me to use a tree data structure to model the expressions

  • Binary - Nodes have at most two children. Informs traversal order.

  • Expression - Means trees represent mathematical expressions that must be evaluated

  • Traversal - Indicates need for structured tree traversal like DFS to visit all nodes

  • Evaluation - I have to calculate expression results at each node by processing children

  • Recursion - Recursive traversal is a natural fit for processing subtrees

  • Operators/Operands - Nodes will be either operators to apply or operands to return

  • Equivalence - Comparing final evaluated results, not just structure, will determine if trees are equivalent

  • Subtree - Smaller nested sub-trees must be evaluated independently

So the core terms like binary expression tree, traversal, evaluation, recursion and equivalence tell me:

  • Use a recursive tree traversal algorithm

  • Compute expression results at each node

  • Track intermediate values in nodes

  • Compare final evaluations to determine equivalency

The key is moving from conceptual terms to specific strategies like recursive DFS for traversal and node-based memoization for evaluation.

We can visualize some key properties and aspects of this problem using tables and diagrams:

Sample Trees:

Tree 1: Tree 2:

       +                    +
     /   \                /   \
   a      +             +      c  
     / \                / \
    b   c              a   b
  • Show example tree structures using standard diagram notation

Node Evaluation:

  • | 9 (2 + 3)
  • Table showing node values during evaluation traversal

Equivalence Check:

Tree 1: 14 Tree 2: 14

  • Simple table comparing final evaluated values

Step-by-step Visualization:

  1. Traverse left subtree
  2. Traverse right subtree
  3. Evaluate + node
  4. Store result in + node
  5. Compare root values (14 == 14)
  • Sequence diagram of incremental steps

These diagrams help visualize the tree structures, evaluation process, equivalence check, and traversal order - both static and step-by-step.

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

Here is one way to break this down into more granular steps:

High-level approach:

  1. Traverse and evaluate each tree
  2. Compare final values

Refined steps:

  1. Build tree data structures from input
  2. Implement recursive traversal function:
    • Check if node is operator or operand
    • If operator, evaluate children and store result
    • If operand, return value
    • Recursively call on children
  3. Traverse tree 1 fully, storing values in nodes
  4. Traverse tree 2 fully, storing values in nodes
  5. Compare root node values of tree 1 and tree 2
  6. Return true if equal, false otherwise

Independent parts:

  • Building tree structures
  • Writing traversal function
  • Comparing final values

Repeatable patterns:

  • Recursively processing child nodes
  • Evaluating and storing node values
  • Comparing two values for equality

This breaks the problem down into modeling the trees, implementing recursive traversal to evaluate nodes, and finally comparing root values. The traversal logic is reusable across both trees.

Solution Approach and Analysis

Here is a step-by-step explanation of how I would approach solving this problem:

  1. Model the input as binary trees
  • Create tree nodes with values and left/right child pointers
  • Build out overall tree structures based on input
  1. Implement depth-first traversal
  • Use recursion to visit nodes systematically
  • Check if node is operator or operand
  1. Evaluate expressions during traversal
  • For operator nodes, recursively evaluate children
  • For operands, simply return value
  • Store result in node
  1. Compare final root values
  • After full traversal, compare root results
  • Return true if equal, false otherwise

This builds up the expression evaluation incrementally using recursive DFS traversal. We compute and cache results in each node until reaching the roots for comparison.

If nodes could have more complex expressions, I would have to handle order of operations carefully. With just addition, the traversal order does not matter.

Example tree:

    /     \                    
  +        3                
 / \        
2   1
  1. Build tree structure

  2. DFS traverse left subtree, 2 + 1 = 3

  3. DFS traverse right, return 3

  4. Multiply: 3 * 3 = 9

  5. Compare root values, 9 == 9 is true

The step-by-step traversal and visualization helps illustrate how we evaluate the overall expression through the individual nodes.

Identify Invariant

An invariant in this problem is a condition or statement that remains true throughout the execution of the solution. Some potential invariants are:

  • The tree structures and node values remain constant. The input trees do not change.

  • The evaluation rules for operators and operands stay the same. How nodes are evaluated is fixed.

  • Once a node is given a value during traversal, that value does not change afterwards.

  • The relative relationships between nodes like child/parent links are preserved. The overall tree topology remains constant.

  • Nodes marked visited during traversal remain visited. Their visited status is not reversed.

  • If two subtrees are equivalent before traversal, they remain equivalent after. Equivalence is invariant.

  • The output variables indicating tree equality start out empty/undefined, and are only set once.

So in essence, the key invariants are:

  • Unchanged input tree structures

  • Consistent, fixed evaluation rules

  • Monotonically computed node values

  • Preserved overall tree topology and relationships

  • Invariant subtree equivalence

The traversal cannot alter these core aspects that define the input space and evaluation logic. The final output also remains unset until traversal and comparison is complete.

Identify Loop Invariant

The loop invariant in this problem is a condition that remains true before and after each iteration of the main traversal loop. Some possibilities for loop invariants are:

  • The input trees remain unchanged by the loop. Their structure and values are the same.

  • The evaluation rules and logic do not change within the loop. Operators and operands are evaluated consistently.

  • A node’s value, once computed, remains the same after each loop iteration.

  • The relative topology and relationships between nodes stays constant across loop iterations.

  • Any subtrees evaluated as equivalent before the loop are still equivalent after each iteration.

  • The output variable indicating overall tree equality remains unset within the loop body.

  • The set of nodes visited only grows with each iteration as more nodes get traversed.

So in summary, the key loop invariants center around the unchanged input trees, consistent node evaluation logic, monotonically computed node values, and preserved topology and equivalences. The loop cannot modify these core aspects that determine the correctness of the overall traversal and comparison.

The invariant refers to high-level conditions that are true for the entire problem, regardless of implementation details.

The loop invariant focuses specifically on what must remain true before and after each iteration of the main traversal loop.

There is overlap - the invariants describing the unchanged tree structures, evaluation rules, node relationships etc. also qualify as loop invariants.

However, some invariants like the final output variable being unset initially are important for the problem but not relevant inside the loop body.

And some loop invariants like the growing set of visited nodes don’t necessarily apply globally, only locally within the loop.

So while there is a strong overlap between the invariant and loop invariant properties in this problem, they are not identical sets. The loop invariants are focused on preserving key attributes directly impacted by the traversal logic itself.

The global invariants provide wider constraints that the full solution must respect, even outside the core traversal loop.

Thought Process

Here is how I would think through solving this problem step-by-step:

The problem statement cues:

  • Binary expression trees suggests using a tree data structure

  • Nodes represent operators and operands which need evaluation

  • We need to compare final evaluation results

  • This tells me we need to traverse and evaluate the trees

The steps would be:

  1. Build tree data structures to represent the input

  2. Implement a recursive traversal function:

    • Check if node is operator or operand
    • For operator, evaluate children and store result in node
    • For operand, simply return value
    • Recursively call function on child nodes
  3. Traverse tree 1 fully, evaluating each node

  4. Traverse tree 2 fully, evaluating each node

  5. Compare root node values - return true if equal

  6. Return false if root values differ

In Python:


class Node:
  def __init__(self, val, left=None, right=None):
    self.val = val
    self.left = left
    self.right = right

def evaluate(node):

  if node.val.isnumeric(): 
    return int(node.val)  

  left_val = evaluate(node.left)
  right_val = evaluate(node.right)
  if node.val == '+':
    return left_val + right_val

  # Other operators   
def isEqual(root1, root2):
  val1 = evaluate(root1)
  val2 = evaluate(root2)

  return val1 == val2

PENDING TEST This evaluates trees by recursion, computing node values incrementally. We compare final root values to determine equivalency.

Establishing Preconditions and Postconditions

Here are the details for the evaluate method:

  1. Parameters:
  • node - The Tree node to evaluate
  • Tree node class type
  • Represents the current node during traversal
  1. Preconditions:
  • Node value is either an operator or operand
  • Node’s children are also valid nodes if present
  • Tree structure is fully built
  1. Method Functionality:
  • Recursively evaluates the expression tree under the given node
  • Computes result based on node’s operator/operand and children
  • Returns the evaluation result for that subtree
  1. Postconditions:
  • Node value fields are populated with evaluation results
  • No other changes to tree structure
  • Returns the computed value for the given subtree
  1. Error Handling:
  • Invalid node value types raise exception
  • Missing children for operator node raises exception
  • Return 0 if invalid node as error state

So in summary, this recursively evaluates a given node’s expression subtree and returns the result. It operates on the node parameters and results in computed value fields.

Problem Decomposition

Here’s how I would break down this problem:

  1. Problem Understanding: Given two expression trees, traverse and evaluate them to numeric values and compare the results to determine if the trees are equivalent.

  2. Initial Breakdown:

  • Parse input into tree structures
  • Implement traversal to evaluate trees
  • Compare final evaluation values
  1. Subproblem Refinement:
  • Build tree nodes and topology
  • Write recursive traversal function
    • Check node type
    • Evaluate operands
    • Compute operator results
  • Traverse tree 1 fully
  • Traverse tree 2 fully
  • Compare root values
  1. Task Identification:
  • Tree construction
  • Node type checking
  • Operand evaluation
  • Operator computation
  • Recursive traversal
  • Root value comparison
  1. Task Abstraction:
  • Tree building
  • Node processing
  • Recursive evaluation
  • Result comparison
  1. Method Naming:
  • build_tree
  • process_node
  • evaluate
  • compare_results
  1. Subproblem Interactions:
  • Build trees initially
  • Implement recursive evaluation logic
  • Traverse trees to compute values
  • Finally, compare roots to determine equivalence

The key stages are modeling the trees, implementing recursive evaluation, fully traversing both trees, and comparing final values. Breaking the evaluation part into reusable node processing tasks is key.

From Brute Force to Optimal Solution

Here is a brute force approach and how we can optimize it:

Brute force:

  1. Generate all possible evaluation orders
  2. Evaluate trees for each order
  3. Compare final values
  4. Return if equal

This tries all possible evaluation orders which is inefficient.

Complexity: Time - O(n!) Space - O(n)


  1. Use DFS traversal instead of permutation generation

This builds up evaluation incrementally.

Complexity: Time - O(n) Space - O(h)

  1. Memoize intermediate results

Avoid recomputing subtrees by storing values.

Complexity: Time - O(n) Space - O(n)

  1. Parallelize subtree evaluations

Child subtrees can be evaluated concurrently.

Complexity: Time - O(log(n)) Space - O(n)

  1. Use hashmap for memoization

Speeds up lookup of computed values.

Complexity: Time - O(n) Space - O(n)

In summary, we optimized from a factoral brute force approach to linear time traversal, added caching to avoid repeated work, parallelized independent work, and improved data structures. This gave a large improvement in time complexity.

Code Explanation and Design Decisions

Here’s how I would explain the reasoning behind this code:

  1. The initial parameters are the root nodes of the two trees. This gives the starting point for traversal and evaluation.

  2. The main loop iterates over the tree nodes recursively. Each iteration represents evaluating the expression subtree under that node.

  3. Branches check node types, directing operators to evaluate children and operands to return their values. This respects the expression structure.

  4. The node value fields are updated to cache intermediate results. This is needed to incrementally build the evaluation.

  5. An invariant is that node relationships remain fixed. This ensures a consistent tree topology and evaluation order.

  6. The final output indicates if the full expression trees, now evaluated, are equivalent. This meets the requirement to compare equivalence of the overall expressions.

In summary, the code traverses the tree performing incremental evaluation, while maintaining the overall tree structure invariantly. The end result satisfies checking if the expression results match to determine equivalency.

Coding Constructs

Here are the high-level explanations for this code:

  1. It uses recursion to incrementally traverse and evaluate expression trees. Also employs caching to avoid repeated work.

  2. The code evaluates complex math expressions step-by-step according to rules, and checks if two expressions end up with the same result.

  3. The logical constructs are a recursive traversal function, processing node values, caching results, and comparing final values.

  4. It recursively simplifies the expression tree by evaluating nodes depth-first, storing results along the way, until reaching the root nodes to compare.

  5. The key steps are processing operator and operand nodes, recursively simplifying subtrees, caching values, and comparing root results. This builds up the overall evaluation.

  6. The core algorithmic strategies are recursive tree traversal for incremental evaluation and memoization for efficiency. This allows methodically simplifying expressions into results.

In summary, the code leverages recursion and caching to evaluate expression trees by incrementally processing nodes in a structured way. The explanations focus on the high-level reasoning rather than language-specific implementation details.

Language Agnostic Coding Drills

Here is how I would break down this code into learning units:

Concepts in increasing difficulty:

  1. Data model definition - Define tree node class with value, left, and right. Basic class creation.

  2. Recursive function - Write function signature for recursive traversal. Key recursion skill.

  3. Base case handling - Check if node is leaf. Foundation of recursion.

  4. Data validation - Differentiate operator vs operand nodes. Important for control flow.

  5. Recursion calls - Evaluate and call function on child nodes. Core of recursion.

  6. Result propagation - Return value from initial call. Allows passing back results.

  7. Subresult caching - Store and reuse node evaluation values. Optimization technique.

  8. Final result comparison - Compare root values. Domain-specific application.

The problem solving approach would involve:

  1. Defining the data model
  2. Writing the recursive skeleton
  3. Adding base case handling
  4. Validating node types
  5. Making recursive calls on children
  6. Propagating values back up
  7. Caching results for efficiency
  8. Comparing final root values

Each coding drill builds up key recursion and tree evaluation skills. We start with the foundations before handling optimization and domain-specific logic.

Targeted Drills in Python

Here are Python coding drills for each concept:

  1. Data model definition
class Node:
  def __init__(self, val, left=None, right=None):
    self.val = val
    self.left = left
    self.right = right
  1. Recursive function
def evaluate(node):

  # Function body
  1. Base case handling
if node is None:
  return 0 # Base case
  1. Data validation
if isinstance(node.val, int):
  # Is operand
elif node.val == '+': 
  # Is operator
  1. Recursion calls
left_val = evaluate(node.left)
right_val = evaluate(node.right)
  1. Result propagation
return left_val + right_val # Return result
  1. Subresult caching
if node.val in cache:
  return cache[node.val] 
  1. Final result comparison
return root1_value == root2_value

We can integrate these by:

  1. Defining Node class
  2. Implementing recursive evaluate()
  3. Adding base case handling
  4. Validating node types
  5. Making recursive calls on children
  6. Caching and returning results
  7. Comparing root values

This gradually builds up the full solution by combining the coding concepts in logical progression.


Similar Problems

Here are 10 related LeetCode problems and why they are similar:

  1. Maximum Depth of Binary Tree - Find depth by recursive traversal, like evaluating expression tree.

  2. Invert Binary Tree - Recursively traverses and updates tree, requiring tracking state.

  3. Validate Binary Search Tree - Traverses tree validating node values, like evaluating nodes.

  4. Serialize and Deserialize Binary Tree - Recursively processes tree into string, requires tracking state.

  5. Same Tree - Compares two trees using recursive traversal, like comparing expressions.

  6. Binary Tree Postorder Traversal - Traverses tree depth-first recursively, like evaluation order.

  7. Lowest Common Ancestor of a Binary Tree - Uses tree traversal to compare parents of two nodes.

  8. Binary Tree Level Order Traversal - Processes tree level-by-level, similar traversal pattern.

  9. Symmetric Tree - Checks if trees are mirrors using recursion, like comparing.

  10. Binary Tree Zigzag Level Order Traversal - Traverses tree recursively tracking state, like expression evaluation.

The common themes are recursive tree processing, tracking state, comparing results, and different traversal orders - all relevant skills for evaluating and comparing expression trees.