# 951. 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]`.

``````class Solution {
public:
bool flipEquiv(TreeNode* root1, TreeNode* root2) {
if (!root1 && !root2) return true;
if (!root1 || !root2 || 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));
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/951

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

https://leetcode.com/problems/flip-equivalent-binary-trees/discuss/218358/Java-Iterative-BFS-easy-to-understand

https://leetcode.com/problems/flip-equivalent-binary-trees/discuss/200514/JavaPython-3-DFS-3-liners-and-BFS-with-explanation-time-and-space%3A-O(n).

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