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.

 

Analyse: the same as Same Tree

1. Recursion

    Runtime: 8ms.

 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         if(!root) return true;
14         return symmetric(root->left, root->right);
15     }
16     bool symmetric(TreeNode* leftNode, TreeNode* rightNode){
17         if(!leftNode && !rightNode) return true;
18         if(!leftNode || !rightNode) return false;
19         
20         return leftNode->val == rightNode->val &&
21                symmetric(leftNode->left, rightNode->right) &&
22                symmetric(leftNode->right, rightNode->left);
23     }
24 };

2. Iteration

    Runtime: 4ms.

 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         if(!root) return true;
14         stack<TreeNode* > stk;
15         stk.push(root->left);
16         stk.push(root->right);
17         
18         while(!stk.empty()){
19             TreeNode* node1 = stk.top();
20             stk.pop();
21             TreeNode* node2 = stk.top();
22             stk.pop();
23             
24             if(!node1 && !node2) continue;
25             if(!node1 || !node2) return false;
26             if(node1->val != node2->val) return false;
27             
28             stk.push(node1->left);
29             stk.push(node2->right);
30             stk.push(node1->right);
31             stk.push(node2->left);
32         }
33         return true;
34     }
35 };

 

 

posted @ 2015-07-28 13:29  amazingzoe  阅读(159)  评论(0)    收藏  举报