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.
 * 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) {
        return dfs(root1, root2);
    }

    bool dfs(TreeNode* root1, TreeNode* root2) {
        if (root1 == root2 || (!root1 && !root2)) return true;
        if (!root1 || !root2 || root1->val != root2->val) return false;
        return (dfs(root1->left, root2->left) && dfs(root1->right, root2->right)) || (dfs(root1->left, root2->right) && dfs(root1->right, root2->left));
    }
};