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 depthfirst 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:


In this code, we use Counter
from collections
to store the counts of variables, and a helper function dfs
to perform the depthfirst 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:
94. Binary Tree Inorder Traversal: You need to return the inorder traversal of a binary tree.
144. Binary Tree Preorder Traversal: This problem requires you to return the preorder traversal of a binary tree.
145. Binary Tree Postorder Traversal: This problem involves the postorder traversal of a binary tree.
150. Evaluate Reverse Polish Notation: You need to evaluate an arithmetic expression given in Reverse Polish Notation, which involves concepts of expression evaluation.
652. Find Duplicate Subtrees: You are required to find duplicate subtrees in a binary tree, which involves tree comparison.
872. LeafSimilar Trees: This problem requires you to compare the leaf sequences of two binary trees.
100. Same Tree: This problem asks whether two binary trees are identical.
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.
993. Cousins in Binary Tree: This problem is about identifying sibling nodes in a binary tree.
958. Check Completeness of a Binary Tree: You are required to check whether a binary tree is complete.
As for the followup 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 highlevel process  build trees, traverse/evaluate, compare values, return true/false result.
Create small example expression trees that are equivalent and nonequivalent. 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:
Given:
 Rooted trees T1 and T2
Determine:
 If eval(T1) = eval(T2)
Return:
 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.
Terminology
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. depthfirst or breadthfirst. 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, depthfirst vs breadthfirst 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
Analogy:
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 stepbystep, 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.
Constraints
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 overengineering 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:
 Identical trees
Input: Tree 1: [+, a, b] Tree 2: [+, a, b]
Output: True
Analysis: Basic case, same structure and values.
 Different leaf values
Input:
Tree 1: [+, a, b]
Tree 2: [+, a, c]
Output: False
Analysis: Different operand values lead to different evaluation.
 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.
 Overflow
Input: Tree 1: [+, 9999, 1] Tree 2: [+, 9998, 2]
Output: False
Analysis: Large values could cause overflow.
 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 multicolor edges to highlight nonequivalent 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 highlevel 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 depthfirst or breadthfirst 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 nontechnical 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 stepbystep, 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 stepbystep explanation of how I would approach solving this problem:
 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
 Implement a tree traversal function
 Use depthfirst search (DFS) to visit all nodes systematically
 Recursively call traversal on child nodes
 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
 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.
Example:
* *
/ \ / \
+ C A +
/ \ / \
A B B C
Build tree structures
Traverse left tree DFS, compute A + B = X, store in * node
Traverse right tree DFS, compute B + C = Y, store in * node
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 stepbystep tree diagrams.
Inference of ProblemSolving 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 subtrees 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 nodebased 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:
Node  Value 

a  2 
b  3 
c  4 
  9 (2 + 3)
 Table showing node values during evaluation traversal
Equivalence Check:
Tree 1: 14 Tree 2: 14
 Simple table comparing final evaluated values
Stepbystep Visualization:
 Traverse left subtree
 Traverse right subtree
 Evaluate + node
 Store result in + node
 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 stepbystep.
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:
Highlevel approach:
 Traverse and evaluate each tree
 Compare final values
Refined steps:
 Build tree data structures from input
 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
 Traverse tree 1 fully, storing values in nodes
 Traverse tree 2 fully, storing values in nodes
 Compare root node values of tree 1 and tree 2
 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 stepbystep explanation of how I would approach solving this problem:
 Model the input as binary trees
 Create tree nodes with values and left/right child pointers
 Build out overall tree structures based on input
 Implement depthfirst traversal
 Use recursion to visit nodes systematically
 Check if node is operator or operand
 Evaluate expressions during traversal
 For operator nodes, recursively evaluate children
 For operands, simply return value
 Store result in node
 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
Build tree structure
DFS traverse left subtree, 2 + 1 = 3
DFS traverse right, return 3
Multiply: 3 * 3 = 9
Compare root values, 9 == 9 is true
The stepbystep 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 highlevel 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 stepbystep:
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:
Build tree data structures to represent the input
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
Traverse tree 1 fully, evaluating each node
Traverse tree 2 fully, evaluating each node
Compare root node values  return true if equal
Return false if root values differ
In Python:


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:
 Parameters:
node
 The Tree node to evaluate Tree node class type
 Represents the current node during traversal
 Preconditions:
 Node value is either an operator or operand
 Node’s children are also valid nodes if present
 Tree structure is fully built
 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
 Postconditions:
 Node value fields are populated with evaluation results
 No other changes to tree structure
 Returns the computed value for the given subtree
 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:
Problem Understanding: Given two expression trees, traverse and evaluate them to numeric values and compare the results to determine if the trees are equivalent.
Initial Breakdown:
 Parse input into tree structures
 Implement traversal to evaluate trees
 Compare final evaluation values
 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
 Task Identification:
 Tree construction
 Node type checking
 Operand evaluation
 Operator computation
 Recursive traversal
 Root value comparison
 Task Abstraction:
 Tree building
 Node processing
 Recursive evaluation
 Result comparison
 Method Naming:
 build_tree
 process_node
 evaluate
 compare_results
 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:
 Generate all possible evaluation orders
 Evaluate trees for each order
 Compare final values
 Return if equal
This tries all possible evaluation orders which is inefficient.
Complexity: Time  O(n!) Space  O(n)
Optimizations:
 Use DFS traversal instead of permutation generation
This builds up evaluation incrementally.
Complexity: Time  O(n) Space  O(h)
 Memoize intermediate results
Avoid recomputing subtrees by storing values.
Complexity: Time  O(n) Space  O(n)
 Parallelize subtree evaluations
Child subtrees can be evaluated concurrently.
Complexity: Time  O(log(n)) Space  O(n)
 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:
The initial parameters are the root nodes of the two trees. This gives the starting point for traversal and evaluation.
The main loop iterates over the tree nodes recursively. Each iteration represents evaluating the expression subtree under that node.
Branches check node types, directing operators to evaluate children and operands to return their values. This respects the expression structure.
The node value fields are updated to cache intermediate results. This is needed to incrementally build the evaluation.
An invariant is that node relationships remain fixed. This ensures a consistent tree topology and evaluation order.
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 highlevel explanations for this code:
It uses recursion to incrementally traverse and evaluate expression trees. Also employs caching to avoid repeated work.
The code evaluates complex math expressions stepbystep according to rules, and checks if two expressions end up with the same result.
The logical constructs are a recursive traversal function, processing node values, caching results, and comparing final values.
It recursively simplifies the expression tree by evaluating nodes depthfirst, storing results along the way, until reaching the root nodes to compare.
The key steps are processing operator and operand nodes, recursively simplifying subtrees, caching values, and comparing root results. This builds up the overall evaluation.
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 highlevel reasoning rather than languagespecific implementation details.
Language Agnostic Coding Drills
Here is how I would break down this code into learning units:
Concepts in increasing difficulty:
Data model definition  Define tree node class with value, left, and right. Basic class creation.
Recursive function  Write function signature for recursive traversal. Key recursion skill.
Base case handling  Check if node is leaf. Foundation of recursion.
Data validation  Differentiate operator vs operand nodes. Important for control flow.
Recursion calls  Evaluate and call function on child nodes. Core of recursion.
Result propagation  Return value from initial call. Allows passing back results.
Subresult caching  Store and reuse node evaluation values. Optimization technique.
Final result comparison  Compare root values. Domainspecific application.
The problem solving approach would involve:
 Defining the data model
 Writing the recursive skeleton
 Adding base case handling
 Validating node types
 Making recursive calls on children
 Propagating values back up
 Caching results for efficiency
 Comparing final root values
Each coding drill builds up key recursion and tree evaluation skills. We start with the foundations before handling optimization and domainspecific logic.
Targeted Drills in Python
Here are Python coding drills for each concept:
 Data model definition


 Recursive function


 Base case handling


 Data validation


 Recursion calls


 Result propagation


 Subresult caching


 Final result comparison


We can integrate these by:
 Defining Node class
 Implementing recursive evaluate()
 Adding base case handling
 Validating node types
 Making recursive calls on children
 Caching and returning results
 Comparing root values
This gradually builds up the full solution by combining the coding concepts in logical progression.
Q&A
Similar Problems
Here are 10 related LeetCode problems and why they are similar:
Maximum Depth of Binary Tree  Find depth by recursive traversal, like evaluating expression tree.
Invert Binary Tree  Recursively traverses and updates tree, requiring tracking state.
Validate Binary Search Tree  Traverses tree validating node values, like evaluating nodes.
Serialize and Deserialize Binary Tree  Recursively processes tree into string, requires tracking state.
Same Tree  Compares two trees using recursive traversal, like comparing expressions.
Binary Tree Postorder Traversal  Traverses tree depthfirst recursively, like evaluation order.
Lowest Common Ancestor of a Binary Tree  Uses tree traversal to compare parents of two nodes.
Binary Tree Level Order Traversal  Processes tree levelbylevel, similar traversal pattern.
Symmetric Tree  Checks if trees are mirrors using recursion, like comparing.
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.