Description
https://leetcode.com/problems/flip-equivalent-binary-trees/
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 equivelent 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
Example 4:
Input: root1 = [0,null,1], root2 = [] Output: false
Example 5:
Input: root1 = [0,null,1], root2 = [0,1] Output: true
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].
Explanation
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: TreeNode, root2: TreeNode) -> bool:
if root1 == root2:
return True
if not root1 or not root2 or 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))- Time Complexity: ~min(N1, N2)
- Space Complexity: ~min(N1, N2)