题目：

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5

代码：

方法一——递归1：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * 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) return NULL;
        if (!head->next) return new TreeNode(head->val);
        ListNode *slow = head, *fast = head, *last = slow;
        while (fast->next && fast->next->next) {
            last = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = slow->next;
        last->next = NULL;
        TreeNode *cur = new TreeNode(slow->val);
        if (head != slow) cur->left = sortedListToBST(head);
        cur->right = sortedListToBST(fast);
        return cur;
    }
};

方法二——递归2：

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

想法：擅于用递归