LintCode: Binary Tree Postorder Traversal

简介:

C++,递归

复制代码
 1 /**
 2  * Definition of TreeNode:
 3  * class TreeNode {
 4  * public:
 5  *     int val;
 6  *     TreeNode *left, *right;
 7  *     TreeNode(int val) {
 8  *         this->val = val;
 9  *         this->left = this->right = NULL;
10  *     }
11  * }
12  */
13 class Solution {
14     /**
15      * @param root: The root of binary tree.
16      * @return: Postorder in vector which contains node values.
17      */
18 public:
19     vector<int> postorderTraversal(TreeNode *root) {
20         // write your code here
21         vector<int> result;
22         if (root == NULL) {
23             return result;
24         }
25         if (root->left != NULL) {
26             vector<int> left = postorderTraversal(root->left);
27             result.reserve(result.size() + left.size());
28             result.insert(result.end(), left.begin(), left.end());
29         }
30         if (root->right != NULL) {
31             vector<int> right = postorderTraversal(root->right);
32             result.reserve(result.size() + right.size());
33             result.insert(result.end(), right.begin(), right.end());
34         }
35         result.push_back(root->val);
36         return result;
37     }
38 };
复制代码

C++,递归,辅助函数

复制代码
 1 /**
 2  * Definition of TreeNode:
 3  * class TreeNode {
 4  * public:
 5  *     int val;
 6  *     TreeNode *left, *right;
 7  *     TreeNode(int val) {
 8  *         this->val = val;
 9  *         this->left = this->right = NULL;
10  *     }
11  * }
12  */
13 class Solution {
14     /**
15      * @param root: The root of binary tree.
16      * @return: Postorder in vector which contains node values.
17      */
18 public:
19     vector<int> postorderTraversal(TreeNode *root) {
20         // write your code here
21         vector<int> result;
22         if (root == NULL) {
23             return result;
24         } else {
25             postorderCore(root, result);
26         }
27         return result;
28     }
29     void postorderCore(TreeNode *root, vector<int> &result) {
30         if (root == NULL) {
31             return;
32         }
33         if (root->left != NULL) {
34             postorderCore(root->left, result);
35         }
36         if (root->right != NULL) {
37             postorderCore(root->right, result);
38         }
39         result.push_back(root->val);
40         return;
41     } 
42 };
复制代码

C++,非递归

[一个stack]

[一个cur指针]

[一个pre指针]

复制代码
 1 /**
 2  * Definition of TreeNode:
 3  * class TreeNode {
 4  * public:
 5  *     int val;
 6  *     TreeNode *left, *right;
 7  *     TreeNode(int val) {
 8  *         this->val = val;
 9  *         this->left = this->right = NULL;
10  *     }
11  * }
12  */
13 class Solution {
14     /**
15      * @param root: The root of binary tree.
16      * @return: Postorder in vector which contains node values.
17      */
18 public:
19     vector<int> postorderTraversal(TreeNode *root) {
20         // write your code here
21         vector<int> result;
22         if (root == NULL) {
23             return result;
24         }
25         
26         TreeNode *cur = root, *pre = NULL;
27         stack<TreeNode *> sta;
28         
29         while (cur != NULL || !sta.empty()) {
30             while (cur != NULL) {
31                 sta.push(cur);
32                 cur = cur->left;
33             }
34             cur = sta.top();
35             if (cur->right == NULL || cur->right == pre) {
36                 sta.pop();
37                 result.push_back(cur->val);
38                 pre = cur;
39                 cur = NULL;
40             } else {
41                 cur = cur->right;
42             }
43         }
44         return result;
45     }
46 };
复制代码

 


本文转自ZH奶酪博客园博客,原文链接:http://www.cnblogs.com/CheeseZH/p/4999458.html,如需转载请自行联系原作者

相关文章
|
11月前
Leetcode Binary Tree Postorder Traversal(面试题推荐)
非递后续归遍历二叉树,肯定得用到栈。先序遍历很好写,但后续遍历就不是那么容易了。 只需要设置个指针pre,指向最后输出的那个节点就行了,只要判断cur指针指向的是上次输出节点的父节点,且cur无其他未遍历的节点,这个时候就把cur节点输出即可,然后更改pre。原理是要遍历当前节点,其所有子节点都必须遍历完,因为肯定是先左后右,所以只需一个指针保持前一次输出的结果即可。
47 0
LeetCode 145. Binary Tree Postorder Traversal
给定一个二叉树,返回它的后序遍历。
65 0
LeetCode 145. Binary Tree Postorder Traversal
[LeetCode] Construct Binary Tree from Inorder and Postorder Traversal
链接:https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/难度:Medium题目:106.
1130 0
|
Java
[LeetCode] Construct Binary Tree from Preorder and Inorder Traversal
链接:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/难度:Medium题目:105.
970 0
[LeetCode] Binary Tree Postorder Traversal
This is a fundamental and yet classic problem. I share my three solutions here: Iterative solution using stack --- O(n) time and O(n) space; Recurs...
860 0
[LeetCode] Binary Tree Inorder Traversal
This is a fundamental and yet classic problem. I share my three solutions here: Iterative solution using stack --- O(n) time and O(n) space; Recurs...
700 0