LeetCode刷题笔记（链表）：convert-sorted-list-to-binary-search-tree

素颜马尾好姑娘i 2022-06-01 01:39 99阅读 0赞

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*  **转载请注明作者和出处：[http://blog.csdn.net/u011475210][http_blog.csdn.net_u011475210]**
 *  **代码地址：[https://github.com/WordZzzz/Note/tree/master/LeetCode][https_github.com_WordZzzz_Note_tree_master_LeetCode]**
 *  **刷题平台：[https://www.nowcoder.com/ta/leetcode][https_www.nowcoder.com_ta_leetcode]**
 *  **题  库：Leetcode经典编程题**
 *  **编  者：WordZzzz**

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*   *  题目描述
     *  解题思路
     *  C版代码实现

## 题目描述 ##

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

## 解题思路 ##

升序链表转换成高度平衡的二叉搜索树，我们只需要找到链表的中点当作root然后左右递归就可以了。求链表中点当然还是用快慢指针了。

## C++版代码实现 ##

/**
     * 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) {
            return toBST(head, NULL);
        }
    
        TreeNode *toBST(ListNode *head, ListNode *tail){
            if(head == tail)
                return NULL;
            ListNode *fast = head;
            ListNode *slow = head;
            while(fast != tail && fast->next != tail){
                slow = slow->next;
                fast = fast->next->next;
            }
            TreeNode *root = new TreeNode(slow->val);
            root->left = toBST(head, slow);
            root->right = toBST(slow->next, tail);
    
            return root;
        }
    };

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[http_blog.csdn.net_u011475210]: http://blog.csdn.net/u011475210
[https_github.com_WordZzzz_Note_tree_master_LeetCode]: https://github.com/WordZzzz/Note/tree/master/LeetCode
[https_www.nowcoder.com_ta_leetcode]: https://www.nowcoder.com/ta/leetcode