Flip Equivalent Binary Trees
<p>For a binary tree <strong>T</strong>, we can define a <strong>flip operation</strong> as follows: choose any node, and swap the left and right child subtrees.</p>

<p>A binary tree <strong>X</strong>&nbsp;is <em>flip equivalent</em> to a binary tree <strong>Y</strong> if and only if we can make <strong>X</strong> equal to <strong>Y</strong> after some number of flip operations.</p>

<p>Given the roots of two binary trees <code>root1</code> and <code>root2</code>, return <code>true</code> if the two trees are flip equivalent or <code>false</code> otherwise.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="Flipped Trees Diagram" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0951.Flip%20Equivalent%20Binary%20Trees/images/tree_ex.png" style="width: 500px; height: 220px;" />
<pre>
<strong>Input:</strong> 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]
<strong>Output:</strong> true
<strong>Explanation: </strong>We flipped at nodes with values 1, 3, and 5.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root1 = [], root2 = []
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root1 = [], root2 = [1]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in each tree is in the range <code>[0, 100]</code>.</li>
	<li>Each tree will have <strong>unique node values</strong> in the range <code>[0, 99]</code>.</li>
</ul>
Tree; Depth-First Search; Binary Tree
# 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: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
        def dfs(root1, root2):
            if root1 == root2 or (root1 is None and root2 is None):
                return True
            if root1 is None or root2 is None or root1.val != root2.val:
                return False
            return (dfs(root1.left, root2.left) and dfs(root1.right, root2.right)) or (
                dfs(root1.left, root2.right) and dfs(root1.right, root2.left)
            )

        return dfs(root1, root2)
