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.

其实挺难的，一次写对更难。 这种sorted array/list to balanced tree的题目就是递归的找中间然后左右继续

linked list不一样的地方是每次找中间有个小trick,然后就是判断是不是leaf得时候，array用的start==end,这里其实start->next==end。这样的好处是不把mid包括进去，方便遍历的时候以mid为左面终点，和mid->next为右树的起点。。。。

/**
 * 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) {
		return helper(head, 0);
	}

	TreeNode* helper(ListNode* head, ListNode* tail){
		if (head==tail)
			return 0;
		if (head->next==tail)
			return new TreeNode(head->val);
		ListNode *tmp=head, *mid=head;
		while(tmp!=tail && tmp->next!=tail){
			mid=mid->next;
			tmp=tmp->next->next;
		}
		TreeNode* root=new TreeNode(mid->val);
		root->left=helper(head,mid);
		root->right=helper(mid->next,tail);
		return root;
	}
};