951. Flip Equivalent Binary Trees（左右交叉走思想）

problem：

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 root1and 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].

思路：只需要判断

root1->left->val = root2->left->val 与 root1->right->val = root2->right->val 同时成立

或

root1->left->val = root2->right->val 与 root1->right->val = root2->left->val （交叉判断）同时成立

 

另外：如果其中一方出现null而另一方没有null，则返回false

如果两方同时出现null，则证明这条路径已经完整走完，且没有出现不相等的值，我们返回true

/**
 * 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 ||root1->val != root2->val)return false;//其中一方出现null，或值不相等
        
        if(flipEquiv(root1->left,root2->left)&&flipEquiv(root1->right,root2->right)||
         flipEquiv(root1->left,root2->right)&&flipEquiv(root1->right,root2->left) )return true;
        else return false;
    }
};