Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSame(TreeNode *p, TreeNode *q)
    {  
        if (p == NULL && q == NULL)  
            return true;  
        else if (p == NULL || q == NULL)  
            return false;  
        return p->val == q->val && isSame(p->left, q->right)
        && isSame(p->right, q->left);  
    }  
    
    bool isSymmetric(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL)
            return true;
        
        return isSame(root, root);
    }
};