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，但是没有想出DFS的算法来。后来想到了一个递归的方法，但是空间和时间都是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，每个递归中修改，需要很强大的想象力，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