Leetcode #1612: Check If Two Expression Trees are Equivalent
In this guide, we solve Leetcode #1612 Check If Two Expression Trees are Equivalent in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A binary expression tree is a kind of binary tree used to represent arithmetic expressions. Each node of a binary expression tree has either zero or two children.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Tree, Depth-First Search, Hash Table, Binary Tree, Counting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: root1 = [x], root2 = [x]
Output: true
Python Solution
# Definition for a binary tree node.
# class Node(object):
# def __init__(self, val=" ", left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def checkEquivalence(self, root1: 'Node', root2: 'Node') -> bool:
def dfs(root, v):
if root is None:
return
if root.val != '+':
cnt[root.val] += v
dfs(root.left, v)
dfs(root.right, v)
cnt = Counter()
dfs(root1, 1)
dfs(root2, -1)
return all(x == 0 for x in cnt.values())
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.