根据单链表构造二叉查找树 Convert Sorted List to Binary Search Tree

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

可参考姐妹问题：根据有序数组构造二叉查找树 Convert Sorted Array to Binary Search Tree

思路：二叉查找树的构造往往伴随着二分查找的过程。这是一个单链表而不是数组。二分查找不好用。就用递归吧。每次找到链表的最中间结点， 作为树根节点，左右两侧作为左右孩子。

/** * 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;    return fun(head, NULL);    }    TreeNode* fun(ListNode *head, ListNode *end)    {    if(head == end)    return NULL;         ListNode *p,*mid;    p = mid = head;    while(p != end && p->next != end)    {    p = p->next->next;    mid = mid->next;    }    TreeNode *m = new TreeNode(mid->val);    m->left = fun(head, mid);    m->right = fun(mid->next, end);    return m;    }};