Question：
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example:
Given the sorted linked list: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ -3 9
/ /
-10 5
Tips：
给定一个有序的单链表，将其转换成一个平衡二叉查找树。
思路：
（1）找到链表的中间位置，作为BST的根节点。方法就是设置两个指针，fast、slow，slow每次走一步，fast每次走两步。当fast先遇到null，slow就指向链表中间节点。
（2）左、右子树的根节点也用相同的方法找到，并作为根节点的左右子树。接下来的结点可用递归来解决。
代码：
public TreeNode sortedListToBST(ListNode head){ if(head==null) return null; return toBST(head,null); } private TreeNode toBST(ListNode head,ListNode tail) { if(head==tail) return null; ListNode fast=head; ListNode slow=head; //循环找到链表中间位置作为根节点。 while(fast!=tail && fast.next!=tail){ slow=slow.next; fast=fast.next.next; } TreeNode root=new TreeNode(slow.val); root.left=toBST(head,slow); root.right=toBST(slow.next,tail); return root; }
