# LeetCode 951. Flip Equivalent Binary Trees

## Description

https://leetcode.com/problems/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.

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 = 
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]`.

## Python Solution

``````# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def flipEquiv(self, root1: TreeNode, root2: TreeNode) -> bool:
if root1 == root2:
return True

if not root1 or not root2 or root1.val != root2.val:
return False

return (self.flipEquiv(root1.left, root2.left) and self.flipEquiv(root1.right, root2.right)) or (self.flipEquiv(root1.left, root2.right) and self.flipEquiv(root1.right, root2.left))``````
• Time Complexity: ~min(N1, N2)
• Space Complexity: ~min(N1, N2)