101. Symmetric Tree

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]

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

But the following 

[1,2,2,null,3,null,3]

1
/ \
2   2
\   \
3    3

判断一棵二叉树是否是镜像的

C++（15ms）：

1 /**
2  * Definition for a binary tree node.
3  * struct TreeNode {
4  *     int val;
5  *     TreeNode *left;
6  *     TreeNode *right;
7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8  * };
9  */
10 class Solution {
11 public:
12     bool isSymmetric(TreeNode* root) {
13         TreeNode* left ;
14         TreeNode* right ;
15         if (!root)
16             return true ;
17         queue<TreeNode*> q1 , q2 ;
18         q1.push(root->left) ;
19         q2.push(root->right) ;
20         while(!q1.empty() && !q2.empty()){
21             left = q1.front() ;
22             q1.pop() ;
23             right = q2.front() ;
24             q2.pop() ;
25
26             if (left == NULL && right == NULL)
27                 continue ;
28             if (left == NULL || right == NULL)
29                 return false ;
30             if (left->val != right->val)
31                 return false ;
32             q1.push(left->left) ;
33             q1.push(left->right) ;
34             q2.push(right->right) ;
35             q2.push(right->left) ;
36         }
37         return true ;
38     }
39 };