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.

循环地解可以用两个队列分别存储left和right的节点，这里用了递归来解。

关键是把isSymmetric变成isSame，然后进行递归求解。

/**
 * Definition for a binary tree node.
 * 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 isSame(root->left, root->right);
    }
    bool isSame(TreeNode * a, TreeNode * b){
        if(a == NULL && b == NULL){
            return true;
        }
        else if(a == NULL || b == NULL){
            return false;
        }
        else{
            if(a->val != b->val){
                return false;
            }
            else{
                return isSame(a->left, b->left) && isSame(a->right, b->right);
            }
        }
    }
};