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

之前还以为可以一分为二，一次遍历搞定，不仅写乱套了，而且今天晚上才发现根本就不是BST啊！！

其实也没多难……递归算下去就OK

/**
 * 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* buildBST(ListNode* head,ListNode* tail) {
        if(head == tail) {
            return NULL;
        }
        if(head->next == tail) {
            TreeNode* t = new TreeNode(head -> val);
            return t;
        }
        ListNode* n = head;
        ListNode* t = head;
        while(t != tail && t->next != tail) {
            n = n -> next;
            t = t -> next -> next;
        }
        TreeNode* m = new TreeNode(n -> val);
        m -> left = buildBST(head , n);
        m -> right = buildBST(n->next ,tail);
        return m;
    }
    TreeNode* sortedListToBST(ListNode* head) {
        return  buildBST(head,NULL);
    }

};