#Leetcode# 951. Flip Equivalent Binary Trees

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.

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.

 

Note:

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

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        if(!root1 && !root2) return true;
        if(!root1 || !root2) return false;
        if(root1 -> val != root2 -> val) return false;
        return flipEquiv(root1 -> left, root2 -> right) && flipEquiv(root1 -> right, root2 -> left) ||
            flipEquiv(root1 -> left, root2 -> left) && flipEquiv(root1 -> right, root2 -> right);
    }
};

　　为什么要  这个判断