【leetcode】Symmetric Tree

Symmetric Tree

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

对称的递归表达式：

testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);

leftNode->left->val==rightNode->right->val

leftNode->right->val==rightNode->left->val

 

 1 class Solution {
 2 
 3 public:
 4 
 5     bool isSymmetric(TreeNode *root) {
 6 
 7         if(root==NULL)
 8 
 9         {
10 
11             return true;
12 
13         }
14 
15         return testMirror(root->left,root->right);
16 
17        
18 
19     }
20 
21    
22 
23     bool testMirror(TreeNode *leftNode,TreeNode *rightNode)
24 
25     {
26 
27        
28 
29         if(leftNode==NULL&&rightNode==NULL)
30 
31             return true;
32 
33         if(leftNode!=NULL&&rightNode==NULL||leftNode==NULL&&rightNode!=NULL)
34 
35             return false;
36 
37         if(leftNode->val!=rightNode->val)
38 
39             return false;
40 
41            
42 
43         return testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);
44 
45        
46 
47        
48 
49     }
50 
51 };

 

 

 

 

 

posted @ 2015-01-26 23:30  H5开发技术  阅读(235)  评论(0)  编辑  收藏  举报