题目描述：Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
即将一个有序链表转换为一个平衡的二叉搜索树，即AVL树
思路：找到这个有序链表的中间结点，从而划分出左右两个部分，将中间结点作为根节点，从而分别建立出这个根结点的左右子树

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode*root=NULL; {
 *     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;
        if(head->next==NULL)
            return new TreeNode(head->val);

        ListNode *slow=head;
        ListNode *fast=head->next->next;
        while(fast && fast->next)
        {
            slow=slow->next;
            fast=fast->next->next;
        }

        ListNode *l=head;
        ListNode *m=slow->next;
        ListNode *r=m->next;
        slow->next=NULL;

        TreeNode*root=new TreeNode(m->val);
        root->left=sortedListToBST(l);
        root->right=sortedListToBST(r);
        return root;
    }
};