Convert Sorted List to Balanced BST

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

Example

               2
1->2->3  =>   / \
             1   3

分析：
非常简单，用递归即可。需要注意返回mid node的时候，要把整个list分成两半。

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) { val = x; }
  * }
  */
 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public TreeNode sortedListToBST(ListNode head) {
         if (head == null) return null;
         ListNode mid = middle(head);
         TreeNode root = new TreeNode(mid.val);
         root.right = sortedListToBST(mid.next);
         if (mid != head) {
             root.left = sortedListToBST(head);
         }
         return root;
     }

     private ListNode middle(ListNode head) {
         if (head == null || head.next == null) return head;
         ListNode pre = null, slow = head, quick = head;

         while(quick.next != null && quick.next.next != null) {
             pre = slow;
             slow = slow.next;
             quick = quick.next.next;
         }

         if (pre != null) {
             pre.next = null;  // cut the list into halves.
         }
         return slow;
     }
 }