题目：

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

解题思路：

注意是让构造平衡二叉搜索树。

每次将链表从中间断开，分成左右两部分。左边部分用来构造左子树，右边部分用来构造右子树。递归进行求解。

代码：

 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         if (head == NULL) return NULL;
22
23         ListNode *slow = head, *fast = head, *pre = NULL;
24         do {
25             fast = fast->next;
26             if (fast == NULL) break;
27             fast = fast->next;
28
29             pre = slow;
30             slow = slow->next;
31         } while (fast != NULL);
32
33         TreeNode * root = new TreeNode(slow->val);
34         if (pre != NULL) {
35             pre->next = NULL;
36             root->left = sortedListToBST(head);
37         }
38         else {
39             root->left = NULL;
40         }
41         TreeNode * right_root = sortedListToBST(slow->next);
42         root->right = right_root;
43         if (pre != NULL) pre->next = slow;
44         return root;
45     }
46 };

转载于:https://www.cnblogs.com/dongguangqing/p/3728160.html

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解题思路：

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