LeetCode OJ

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：

/**
 * 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) {
        if (head == NULL) return NULL;

        ListNode *slow = head, *fast = head, *pre = NULL;
        do {
            fast = fast->next;
            if (fast == NULL) break;
            fast = fast->next;
            
            pre = slow;
            slow = slow->next;
        } while (fast != NULL);

        TreeNode * root = new TreeNode(slow->val);
        if (pre != NULL) {
            pre->next = NULL;
            root->left = sortedListToBST(head);
        }
        else {
            root->left = NULL;
        }
        TreeNode * right_root = sortedListToBST(slow->next);
        root->right = right_root;
        if (pre != NULL) pre->next = slow;
        return root;
    }
};