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 a binary tree node.
struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

//递归操作。推断一棵二叉树是否对称
class Solution {
public:
	bool isSymmetric(TreeNode* root) {
		if (root == NULL) return true;
		else
		{
				return ifSymmetric(root->left,root->right);
		}
	}

	bool ifSymmetric(TreeNode* tree1, TreeNode* tree2){
		if (tree1==NULL && tree2==NULL) 
			return true;
		else if (tree1 == NULL || tree2 == NULL) 
			return false;
	   
		if (tree1->val != tree2->val) 
			return false;
		else
		{
			return ifSymmetric(tree1->left, tree2->right) && ifSymmetric(tree1->right, tree2->left);
		}
	}
};