109. Convert Sorted List to Binary Search Tree

最新推荐文章于 2020-04-13 06:45:19 发布

ShenYounger

最新推荐文章于 2020-04-13 06:45:19 发布

阅读量288

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/wusecaiyun/article/details/47380751

leetcode 专栏收录该内容

220 篇文章

订阅专栏

本文介绍了一种将升序链表转换为高度平衡二叉搜索树的方法。利用快慢指针定位链表中点，递归构建平衡二叉树。

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

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(head == NULL)return NULL;
        ListNode *fast = head, *slow = head, *preSlow = NULL;
        while(fast->next && fast->next->next)
        {
            fast = fast->next->next;
            preSlow = slow;
            slow = slow->next;
        }
        TreeNode *res = new TreeNode(slow->val);
        fast = slow->next;
        delete slow;
        if(preSlow != NULL)
        {
            preSlow->next = NULL;
            res->left = sortedListToBST(head);
        }
        res->right = sortedListToBST(fast);
        return res;
    }
};