951. Flip Equivalent Binary Trees ------M

 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].

A:

递归比较

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        if(not root1 && not root2)
            return true;
        if(not root1 || not root2)
            return false;
        if(root1->val != root2->val)
            return false;
        int res1 = flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right);
        return res1 || (flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left));
    }
};