LeetCode:Convert Sorted Array to Binary Search Tree,Convert Sorted List to Binary Search Tree

最新推荐文章于 2025-08-15 16:44:13 发布

转载最新推荐文章于 2025-08-15 16:44:13 发布 · 53 阅读

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原文链接：http://www.cnblogs.com/TenosDoIt/p/3440079.html

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#数据结构与算法

本文介绍了如何将已排序的数组和链表转换为高度平衡的二叉搜索树（BST）。通过递归方法，首先找到中间元素作为根节点，然后分别对左右子树进行同样的操作，确保生成的BST高度平衡。

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LeetCode:Convert Sorted Array to Binary Search Tree

Given an array where elements are sorted in ascending order, convert it to a height balanced BST

分析：找到数组的中间数据作为根节点，小于中间数据的数组来构造作为左子树，大于中间数据的数组来构造右子树，递归解法如下

 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     TreeNode *sortedArrayToBST(vector<int> &num) {
13         // IMPORTANT: Please reset any member data you declared, as
14         // the same Solution instance will be reused for each test case.
15         int len = num.size();
16         if(len == 0)return NULL;
17         return sortedArrayToBSTRecur(num, 0, len-1);
18     }
19     TreeNode *sortedArrayToBSTRecur(vector<int> &num, int istart, int iend)
20     {
21         if(istart > iend)return NULL;
22         int middle = (istart+iend)/2;
23         TreeNode *res = new TreeNode(num[middle]);
24         res->left = sortedArrayToBSTRecur(num, istart, middle-1);
25         res->right = sortedArrayToBSTRecur(num, middle+1, iend);
26         return res;
27     }
28 };

LeetCode:Convert Sorted List to Binary Search Tree

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST

分析：和上一题同理，只不过要使用快慢指针来找到链表的中间节点本文地址

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 /**
10  * Definition for binary tree
11  * struct TreeNode {
12  *     int val;
13  *     TreeNode *left;
14  *     TreeNode *right;
15  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
16  * };
17  */
18 class Solution {
19 public:
20     TreeNode *sortedListToBST(ListNode *head) {
21         // IMPORTANT: Please reset any member data you declared, as
22         // the same Solution instance will be reused for each test case.
23         if(head == NULL)return NULL;
24         ListNode *fast = head, *slow = head, *preSlow = NULL;
25         while(fast->next && fast->next->next)
26         {
27             fast = fast->next->next;
28             preSlow = slow;
29             slow = slow->next;
30         }
31         TreeNode *res = new TreeNode(slow->val);
32         fast = slow->next;
33         delete slow;
34         if(preSlow != NULL)
35         {
36             preSlow->next = NULL;
37             res->left = sortedListToBST(head);
38         }
39         res->right = sortedListToBST(fast);
40         return res;
41     }
42 };

转载于:https://www.cnblogs.com/TenosDoIt/p/3440079.html