LeetCode OJ Symmetric Tree 判断是否为对称树（AC代码）

 1 /**
 2  * Definition for binary tree
 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         if(!root)    
14             return true;    //无根结点
15         else if(root->left==(void*)0 && root->right==(void*)0)        //只有根结点的树
16             return true;
17         else if(root->left!=(void*)0 && root->right!=(void*)0 && isSymmetric(root->left,root->right))    //两个子树非空,左右子树为对称
18             return true;
19         else
20             return false;
21     }
22     bool isSymmetric(TreeNode *left,TreeNode *right) {
23         if(left->val==right->val){        //左右结点值相等
24             if(left->left==(void*)0 && right->right==(void*)0 || left->left!=(void*)0 && right->right!=(void*)0 && isSymmetric(left->left,right->right)){        //左右子树都为空、或者对称
25                 if(left->right==(void*)0 && right->left==(void*)0 ||left->right!=(void*)0 && right->left!=(void*)0 && isSymmetric(left->right,right->left))      //右左子树都为空、或对称
26                     return true;
27                 else
28                     return false;
29             }
30             else    //左右子树不对称
31                 return false;
32         }
33         else //左右结点值不等
34             return false;
35     }
36 };