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LeetCode – Refresh – Binary Tree Pre Order Traversal

时间:2015-03-18 08:58:05      阅读:131      评论:0      收藏:0      [点我收藏+]

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Same with inorder.

1. recursive:

 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     void getTree(TreeNode *root, vector<int> &result) {
13         if (!root) return;
14         result.push_back(root->val);
15         getTree(root->left, result);
16         getTree(root->right, result);
17     }
18     vector<int> preorderTraversal(TreeNode *root) {
19         vector<int> result;
20         getTree(root, result);
21         return result;
22     }
23 };

 

2. directly use stack

 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     vector<int> preorderTraversal(TreeNode *root) {
13         vector<int> result;
14         if (!root) return result;
15         stack<TreeNode *> s;
16         while (root || !s.empty()) {
17             if (root) {
18                 s.push(root->right);
19                 result.push_back(root->val);
20                 root = root->left;
21             } else {
22                 root = s.top();
23                 s.pop();
24             }
25         }
26         return result;
27     }
28 };

 

3. two pointers:

 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     vector<int> preorderTraversal(TreeNode *root) {
13         vector<int> result;
14         if (!root) return result;
15         TreeNode *prev = NULL;
16         while (root) {
17             result.push_back(root->val);
18             if (!root->left) {
19                 root = root->right;
20             } else {
21                 prev = root->left;
22                 while(prev->right && prev->right != root) prev = prev->right;
23                 if (!prev->right) {
24                     prev->right = root;
25                     root = root->left;
26                 } else {
27                     prev->right = NULL;
28                     root = root->right;
29                     result.pop_back();
30                 }
31             }
32         }
33         return result;
34     }
35 };

 

LeetCode – Refresh – Binary Tree Pre Order Traversal

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原文地址:http://www.cnblogs.com/shuashuashua/p/4346193.html

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