LeetCode-- Convert Sorted List to Binary Search Tree

最新推荐文章于 2021-11-02 12:48:41 发布

原创最新推荐文章于 2021-11-02 12:48:41 发布 · 478 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#搜索 #链表 #二叉树 #中序遍历 #回溯

leetcode 同时被 2 个专栏收录

89 篇文章

订阅专栏

LeetCode代码思路

89 篇文章

订阅专栏

介绍如何将一个按升序排列的单链表转换为高度平衡的二叉搜索树。通过递归方式实现，确保每个节点的左右子树深度相差不超过1，给出具体算法实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

思路：递归。这道题比上一题难度大，是因为链表无法像数组那样按照索引遍历，那么就需要自底向上的回溯，按照中序遍历的顺序，先确定左字树节点，然后确定根节点，最后确定右子树。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        int len=0;
        ListNode *p=head;
        while(p){
            len++;
            p=p->next;
        }
        return sortedListToBST(head,0,len-1);
    }
    TreeNode* sortedListToBST(ListNode* &list,int start,int end){
        if(start>end) return NULL;
        int mid=(start+end)/2;
        TreeNode *left=sortedListToBST(list,start,mid-1);
        TreeNode *parent=new TreeNode(list->val);
        parent->left=left;
        list=list->next;
        parent->right=sortedListToBST(list,mid+1,end);   
        return parent;
    }
};