<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 {
public:
    bool isSymmetric(TreeNode* root) {
        TreeNode* swapped=new TreeNode();
        if(root==nullptr){
            return root;
        }
        swapped->val=root->val;
        swaptree(root, swapped);

        bool ans;
        ans=issame(root,swapped);
        return ans;
    }

    void swaptree(TreeNode* root, TreeNode* swapped){
        if(root!=nullptr){
            if(root->left!=nullptr){
                swapped->right=new TreeNode(root->left->val);
            }
            if(root->right!=nullptr){
                swapped->left=new TreeNode(root->right->val);
            }
            swaptree(root->left,swapped->right);
            swaptree(root->right,swapped->left);
        }
    }

    bool issame(TreeNode* p, TreeNode* q){
        bool ans=true;
        if(p!=nullptr && q==nullptr){
            return false;
        }
        else if(p==nullptr && q!=nullptr){
            return false;
        }
        if(p!=nullptr && q!=nullptr){
            if(p->val!=q->val){
                return false;
            }
            ans=issame(p->left,q->left);
            if(ans==false){
                return ans;
            }
            ans=issame(p->right,q->right);
            if(ans==false){
                return ans;
            }
        }
        return ans;
    }
};
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