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.

思路：jame bond 的思路，找mid的前一个点，然后两边分开找；这里只传递一个参数；注意middle的前后都要断开，否则会死循环；

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if(head == null) {
            return null;
        }
        ListNode dummpy = new ListNode(-1);
        dummpy.next = head;
        
        ListNode slow = dummpy;
        ListNode fast = dummpy;
        
        // find james bond point;
        while(fast != null && fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        
        ListNode middle = slow.next;
        slow.next = null;
        
        TreeNode root = new TreeNode(middle.val);
        root.left = sortedListToBST(dummpy.next);
        root.right = sortedListToBST(middle.next);
        return root;
    }
}