[LintCode] 二叉树的中序遍历

最新推荐文章于 2025-09-10 15:24:29 发布

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4607756.html

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#数据结构与算法

本文介绍两种二叉树中序遍历的方法：一种使用栈实现，时间复杂度和空间复杂度均为O(n)；另一种采用Morris遍历算法，时间复杂度为O(n)，但仅需O(1)的空间复杂度。

The recursive solution is trivial and I omit it here.

Iterative Solution using Stack (O(n) time and O(n) space):

 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: Inorder in vector which contains node values.
17      */
18 public:
19     vector<int> inorderTraversal(TreeNode *root) {
20         // write your code here
21         vector<int> nodes;
22         TreeNode* node = root;
23         stack<TreeNode*> toVisit;
24         while (node || !toVisit.empty()) {
25             if (node) {
26                 toVisit.push(node);
27                 node = node -> left;
28             }
29             else {
30                 node = toVisit.top();
31                 toVisit.pop();
32                 nodes.push_back(node -> val);
33                 node = node -> right;
34             }
35         }
36         return nodes;
37     }
38 };

Another more sophisticated soltuion using Morris Traversal (O(n) time and O(1) space):

 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: Inorder in vector which contains node values.
17      */
18 public:
19     vector<int> inorderTraversal(TreeNode *root) {
20         // write your code here
21         vector<int> nodes;
22         TreeNode* node = root;
23         while (node) {
24             if (node -> left) {
25                 TreeNode* predecessor = node -> left;
26                 while (predecessor -> right && predecessor -> right != node)
27                     predecessor = predecessor -> right;
28                 if (!(predecessor -> right)) {
29                     predecessor -> right = node;
30                     node = node -> left;
31                 }
32                 else {
33                     predecessor -> right = NULL;
34                     nodes.push_back(node -> val);
35                     node = node -> right;
36                 }
37             }
38             else {
39                 nodes.push_back(node -> val);
40                 node = node -> right;
41             }
42         }
43         return nodes;
44     }
45 };