[Leetcode] 0951. Flip Equivalent Binary Trees

Explore diverse LeetCode solutions in Python, C++, JavaScript, SQL, and TypeScript. Ideal for interview prep, learning, and code practice in multiple programming languages. Github Repo Link

Table of contents

Medium

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.

Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or false otherwise.

Example 1:

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.

Example 2:

Input: root1 = [], root2 = []
Output: true

Example 3:

Input: root1 = [], root2 = [1]
Output: false

Constraints:

The number of nodes in each tree is in the range [0, 100].
Each tree will have unique node values in the range [0, 99].

Python

				
					# time complexity: O(n) # space complexity: O(n) from typing import Optional class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def traverse(node): if node is None: return if node.val: print(node.val) if node.left: traverse(node.left) if node.right: traverse(node.right) class Solution: def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool: if not root1 and not root2: return True if not root1 or not root2: return False if root1.val != root2.val: return False return (self.flipEquiv(root1.left, root2.left) and self.flipEquiv(root1.right, root2.right)) or (self.flipEquiv(root1.left, root2.right) and self.flipEquiv(root1.right, root2.left)) root1 = TreeNode(1) root1.left = TreeNode(2) root1.right = TreeNode(3) root1.left.left = TreeNode(4) root1.left.right = TreeNode(5) root1.right.left = TreeNode(6) root1.left.right.left = TreeNode(7) root1.left.right.right = TreeNode(8) root2 = TreeNode(1) root2.left = TreeNode(3) root2.right = TreeNode(2) root2.left.right = TreeNode(6) root2.right.left = TreeNode(4) root2.right.right = TreeNode(5) root2.right.right.left = TreeNode(8) root2.right.right.right = TreeNode(7) print(Solution().flipEquiv(root1, root2))

Medium

Python

related articles