<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://assets.leetcode.com/uploads/2021/02/19/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://assets.leetcode.com/uploads/2021/02/19/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?
/**
 * 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 {
private:
    // bool isSymmetricHelp(TreeNode* left , TreeNode* right)
    // {
    //     if(left == nullptr && right == nullptr) return left == right;
    //     if(left->val != right->val) return false;
    //     return isSymmetricHelp(left->left,right->right) && isSymmetricHelp(left->right,right->left);
    // }
public:
    bool isSymmetric(TreeNode* root) {
        if(root == nullptr) return true;
        queue<TreeNode*> q;
        q.push(root->left);
        q.push(root->right);
        while(q.size())
        {
            TreeNode* left = q.front();q.pop();
            TreeNode* right = q.front();q.pop();
            if(left == nullptr && right == nullptr) continue;
            else if(left == nullptr || right == nullptr) return left == right;
            if(left->val != right->val) return false;
            q.push(left->left);
            q.push(right->right);
            q.push(left->right);
            q.push(right->left);
        }
        return true;
        //return root == nullptr || isSymmetricHelp(root->left,root->right);
    }
};
17800