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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * 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 isSymmetric(TreeNode *root) 
    {
        if(root == NULL)
            return true;
        return isSymmetric(root->left,root->right);
    }
    bool isSymmetric(TreeNode* left,TreeNode* right)
    {
        if(left==NULL && right==NULL)
            return true;
        if(left==NULL || right==NULL)
            return false;
        bool l = isSymmetric(left->right,right->left);
        if(left->val != right->val)
            return false;
        bool r = isSymmetric(left->left,right->right);
        return l&&r;
    }
};

LeetCode--Symmetric Tree