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

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字数统计: 303 | 阅读时长 ≈ 2

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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

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/**
 * 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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