## 109. 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.

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
``````

Thoughts:

use slow, fast two pointers to divide the tree

Follow up : inplace conversion, assume both use class Node.

``````/**
* 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; }
* }
*/
// try inplace
class Solution {
if (head == null) return null;
}
private TreeNode binaryTraversal(ListNode head, ListNode tail){
while(fast != tail &&  fast.next != tail){
fast = fast.next.next;
slow = slow.next;
}