（6）convert-sorted-list-to-binary-search-tree
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
思路：
（1）分析题意：要求将一条有序链表转化为一颗二叉搜索树
（2）快慢指针法找到链表的中点
（3）以中点的值建立一个树的根节点
（4）从中间节点将原链表分为两条子链表
（5）分别对两个链表递归调用原函数

代码实现：

class Solution {
public:
    TreeNode *sortedListToBST(ListNode *head) {
        if(head == NULL)
            return NULL;
        ListNode* fast = head;
        ListNode* slow = head;
        ListNode* ret = NULL;
        while(fast!=NULL&&fast->next!=NULL)
        {
            fast = fast->next->next;
            ret = slow;
            slow = slow->next;
        }
        TreeNode* root = new TreeNode(slow->val);//找到中间节点作为根节点
        if(ret != NULL)
        {
            ret->next = NULL;
            root->left = sortedListToBST(head);
        }
        if(slow->next != NULL)
        {
            root->right = sortedListToBST(slow->next);
        }
        return root;
    }
};