Flip The Tree!!!!

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

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].

It is actually pretty simple (recursively), 3-lines, cheers, ACC.

public class Solution
{
public bool FlipEquiv(TreeNode root1, TreeNode root2)
{
if (root1 == null && root2 == null) return true;
if (root1 == null || root2 == null || 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));
}
}