Leetcode #951: Flip Equivalent Binary Trees
In this guide, we solve Leetcode #951 Flip Equivalent Binary Trees 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
For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees. A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.
Python Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
def dfs(root1, root2):
if root1 == root2 or (root1 is None and root2 is None):
return True
if root1 is None or root2 is None or root1.val != root2.val:
return False
return (dfs(root1.left, root2.left) and dfs(root1.right, root2.right)) or (
dfs(root1.left, root2.right) and dfs(root1.right, root2.left)
)
return dfs(root1, root2)
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.