951. Flip Equivalent Binary Trees**

951. Flip Equivalent Binary Trees**

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

题目描述

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.

Note:

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

C++ 实现 1

其中

if (!root1 || !root2) return root1 == root2;

不必多说. 在两个 root 均不为空的情况下, 如果它们的值不相等, 那么就返回 false.

而如果它们的值相等, 就需要判断左右子树是不是相等, 或者翻转后左右子树是否相等.

public:
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
     
        if (!root1 || !root2) return root1 == root2;
        if (root1->val != root2->val) return false;
        return (flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right)) ||
            (flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left));
    }
};