Flatten Binary Tree to Linked List

最新推荐文章于 2025-09-10 21:59:39 发布

转载最新推荐文章于 2025-09-10 21:59:39 发布 · 53 阅读

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原文链接：http://www.cnblogs.com/amazingzoe/p/4683466.html

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本文介绍了一种将二叉树结构通过先序遍历的方式展平为链表的方法，并提供了一个运行时间为12毫秒的C++实现方案。展平后的链表中每个节点的右子节点指向先序遍历的下一个节点。

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

The flattened tree should look like:

click to show hints.

Hints:

If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

Analyse: The sequence of the linked list is the preorder traversal. So we can first do the preorder traversal then create a new tree.

Runtime: 12ms.

 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     void flatten(TreeNode* root) {
13         if(!root) return;
14         vector<TreeNode* > vec;
15         preorder(root, vec);
16         
17         TreeNode* current = root;
18         int index = 1;
19         while(index < vec.size()){
20             TreeNode* temp(vec[index]);
21             current->right = temp;
22             current->left = NULL;
23             current = temp;
24             index++;
25         }
26         return;
27     }
28     void preorder(TreeNode* root, vector<TreeNode* >& vec){
29         if(!root) return;
30         
31         vec.push_back(root);
32         preorder(root->left, vec);
33         preorder(root->right, vec);
34     }
35 };