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：怎么获得平衡搜索二叉树呢？取链表的中点作为二叉树的根节点，把链表分为左右两部分，递归调用。

开始我的思路是正确的，但是直接在函数主体中递归调用出了问题，应该定义一个辅助函数实现递归调用的。

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;        TreeNode *root = NULL;        if(head->next==NULL){//链表只有一个值时，用这一个值构建二叉树然后直接返回            TreeNode *node = new TreeNode(head->val);            root = node;            return root;        }        insertnode(head,root);        return root;                }    void insertnode(ListNode *head,TreeNode* &root){        ListNode *slow = head;        ListNode *fast = head;        ListNode *pre = head;//用来记录slow的前一个节点        while(fast!=NULL && fast->next!=NULL){            pre = slow;//先给pre赋值实现 记录slow的前一个节点            slow = slow->next;            fast = fast->next->next;        }        TreeNode *node = new TreeNode(slow->val);        root = node;        pre->next = NULL;//把链表从pre后面截断        root->left = sortedListToBST(head);        root->right = sortedListToBST(slow->next);    }};