Symmetric Tree
<p>Given the <code>root</code> of a binary tree, <em>check whether it is a mirror of itself</em> (i.e., symmetric around its center).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0101.Symmetric%20Tree/images/symtree1.jpg" style="width: 354px; height: 291px;" />
<pre>
<strong>Input:</strong> root = [1,2,2,3,4,4,3]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0101.Symmetric%20Tree/images/symtree2.jpg" style="width: 308px; height: 258px;" />
<pre>
<strong>Input:</strong> root = [1,2,2,null,3,null,3]
<strong>Output:</strong> false
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you solve it both recursively and iteratively?
Tree; Depth-First Search; Breadth-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 isSymmetric(TreeNode* root) {
        auto dfs = [&](this auto&& dfs, TreeNode* root1, TreeNode* root2) -> bool {
            if (root1 == root2) {
                return true;
            }
            if (!root1 || !root2 || root1->val != root2->val) {
                return false;
            }
            return dfs(root1->left, root2->right) && dfs(root1->right, root2->left);
        };
        return dfs(root->left, root->right);
    }
};
