0951._Flip_Equivalent_Binary_Trees.md

951. Flip Equivalent Binary Trees

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/flip-equivalent-binary-trees/

内容描述

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.

Write a function that determines whether two binary trees are flip equivalent.  The trees are given by root nodes root1 and root2.

 

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.
Flipped Trees Diagram
 

Note:

Each tree will have at most 100 nodes.
Each value in each tree will be a unique integer in the range [0, 99].

解题方案

思路 1 - 时间复杂度: O(N)- 空间复杂度: O(1)******

递归，首先root1和root2肯定要相等才行，然后要么左左相等且右右相等，要么左右相等且右左相等

class Solution:
    def flipEquiv(self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        """
        if not root1 and not root2:
            return True
        elif not (root1 and root2):
            return False
        else:
            if root1.val == root2.val:
                return (self.flipEquiv(root1.left, root2.right) and self.flipEquiv(root1.right, root2.left)) or (self.flipEquiv(root1.left, root2.left) and self.flipEquiv(root1.right, root2.right))
            else:
                return False