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.
想了好久想不出来。后来看了题目分类里面说是DFS，可是没有想出DFS的算法来。后来想到了一个递归的方法。可是空间和时间都是O(n)。
后来网上找了一个空间是O(1)的时间是O(n)的算法，是一种新的解题思路，用的是中递归。

一般我解题都是用的头递归或者尾递归。第一次见识到了中递归，相当于把递归当成了一个循环体，用引用来作为变量，每一个递归中改动，须要非常强大的想象力。把整个递归树在脑子里想清楚。

空间和时间都为O(n):

    TreeNode *sortedListToBST(ListNode *head) 
	{
		vector<TreeNode*> treeNodes;
		while (head != NULL)
		{
			TreeNode *node = new TreeNode(head->val);
			treeNodes.push_back(node);
			head = head->next;
		}
		return genBST(0, treeNodes.size()-1, treeNodes);
    }
	
	TreeNode* genBST(int start, int end, vector<TreeNode*> &treeNodes)
	{
		if (start == end) return treeNodes[start];
		else if (start+1 == end)
		{
			treeNodes[start]->right = treeNodes[end];
			return treeNodes[start];
		} 

		int mid = (start+end)/2;
		TreeNode* root = treeNodes[mid];
		root->left = genBST(start, mid-1, treeNodes);
		root->right = genBST(mid+1, end, treeNodes);
		return root;
	}

空间为O(1)时间为O(n)：

	TreeNode *sortedListToBST(ListNode *head)
	{
	    int len = 0;
        ListNode * node = head;
        while (node != NULL)
        {
            node = node->next;
            len++;
        }
        return buildTree(head, 0, len-1);
    }
    
    TreeNode *buildTree(ListNode *&node, int start, int end)
    {
        if (start > end) return NULL;
        int mid = start + (end - start)/2;
        TreeNode *left = buildTree(node, start, mid-1);
        TreeNode *root = new TreeNode(node->val);
        root->left = left;
        node = node->next;
        root->right = buildTree(node, mid+1, end);
        return root;
    }

解法引用：http://www.bwscitech.com/a/jishuzixun/javayuyan/2013/0930/15822.html