题目：

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

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

But the following [1,2,2,null,3,null,3] is not:

1 / \ 2 2 \ \ 3 3

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

思路：

这道题目，其实很简单，只要稍稍修改我前一篇博客的函数，并调用就好了Same Tree

代码：

/**
 * 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 isSameTree(TreeNode* p, TreeNode* q) {  
        if(p==NULL&&q==NULL)  
            return true;  
        else if(p==NULL&&q!=NULL)  
            return false;  
        else if(p!=NULL&&q==NULL)  
            return false;  
        else  
        {  
            if(p->val!=q->val)  
                return false;  
            else  
                return isSameTree(p->left,q->right)&&isSameTree(p->right,q->left);  
        }  
          
    }  
    bool isSymmetric(TreeNode* root) {
        TreeNode* temp = root;
        return isSameTree(root,temp);       
    }
};