Binary Tree Inorder Traversal

递归与迭代实现二叉树中序遍历的算法对比

最新推荐文章于 2025-09-09 22:10:29 发布

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最新推荐文章于 2025-09-09 22:10:29 发布

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文章标签： runtime

原文链接：http://www.cnblogs.com/amazingzoe/p/4679898.html

本文详细介绍了如何使用递归和迭代两种方法实现二叉树的中序遍历，并通过实例代码进行验证。重点讨论了两种方法的时间复杂度、空间复杂度以及各自的优缺点，旨在帮助读者理解不同场景下选择合适遍历方式的重要性。

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

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

Analyse: root->left, root, root->right.

1. Recursion

Runtime: 0ms.

 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     vector<int> inorderTraversal(TreeNode* root) {
13         vector<int> result;
14         if(!root) return result;
15         
16         inorder(root, result);
17         return result;
18     }
19     void inorder(TreeNode* root, vector<int>& result){
20         if(root->left) inorder(root->left, result);
21         result.push_back(root->val);
22         if(root->right) inorder(root->right, result);
23     }
24 };

2. Iteration

Runtime: 0ms.

 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     vector<int> inorderTraversal(TreeNode* root) {
13         vector<int> result;
14         if(!root) return result;
15         stack<TreeNode* > stk;
16         
17         while(root || !stk.empty()){
18             if(root){
19                 stk.push(root);
20                 root = root->left;
21             }
22             else{
23                 root = stk.top();
24                 result.push_back(root->val);
25                 stk.pop();
26                 root = root->right;
27             }
28         }
29         return result;
30     }
31 };