# Flip Equivalent Binary Trees Problem

## Description

LeetCode Problem 951.

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 Xis 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 equivalent or false otherwise.

Example 1:

``````1
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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:

``````1
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Input: root1 = [], root2 = []
Output: true
``````

Example 3:

``````1
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Input: root1 = [], root2 = [1]
Output: false
``````

Example 4:

``````1
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Input: root1 = [0,null,1], root2 = []
Output: false
``````

Example 5:

``````1
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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].

## Sample C++ Code

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/**
* 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 (root1 == NULL and root2 == NULL)
return true;
if (root1 == NULL or root2 == NULL)
return false;
if (root1->val != root2->val)
return false;

bool non_flip = flipEquiv(root1->left, root2->left) &&
flipEquiv(root1->right, root2->right);
bool flip = flipEquiv(root1->left, root2->right) &&
flipEquiv(root1->right, root2->left);

return non_flip || flip;
}
};
``````