Description

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].

分析

题目的意思是：给出两个二叉树，问两个二叉树能够通过左右变换使得二叉树相等。这道题用递归的方法就是如果当前节点不相等就直接返回Fasle，如果相等的话，其左右分支右两种情况满足条件都可以，第一种是左右=左右；另一种是左右=右左。然后写一下递归的终止条件就行了。如果能够想到就很简单。

代码

# 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 is None and root2 is None):
            return True
        if(root1 is None or root2 is None):
            return False
        if(root1.val==root2.val):
            left=self.flipEquiv(root1.left,root2.left)
            right=self.flipEquiv(root1.right,root2.right)
            if(left and right):
                return True
            left=self.flipEquiv(root1.right,root2.left)
            right=self.flipEquiv(root1.left,root2.right)
            if(left and right):
                return True
            else:
                return False
        else:
            return False