[LeetCode]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.

由于链表不能随机访问，只能修改链表结构，要不就只能遍历求长度，再求出中间位置。

这里不能修改结构，总是要不断的遍历，求中间位置，十分罗嗦。

简单如前一题，遍历一遍，把值 push_back 到 vector<int> num 里，同前一题。

/**
 * 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 *help(vector<int> &num,int a,int b){
		int len  = b - a + 1;
		if(len <= 0 )
			return NULL;
		int mid  = a+len/2;
		TreeNode *p = new TreeNode(num[mid]);
		if(len ==1)
			return p;
		int left_len = mid - a;
		int right_len = b - mid;
		if(left_len > 0)
			p->left = help(num,a,a+left_len-1);
		if(right_len > 0)
			p->right = help(num,a+left_len+1, b);
		return p;
	}
    TreeNode *sortedListToBST(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
		vector<int> num;
		num.clear();
		while(head)
		{
			num.push_back(head->val);
			head = head->next;
		}
           int len = num.size();
		if(len==0)
			return NULL;
		return help(num,0,len-1);
    }
};