Convert Sorted List to Binary Search Tree.java

/*
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


time = O(nlogn) for each find middle O(n), logn heads
space = O(logn)
*/


/**
 * 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; }
 * }
 */
class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if (head == null) {
            return null;
        }
        return bst(head, null);
    }
    
    private TreeNode bst(ListNode head, ListNode end) {
        ListNode slow = head;
        ListNode fast = head;
        if (head == end) {
            return null;
        }
        // this is different from normal mid
        while (fast.next != end && fast.next.next != end) {
            slow = slow.next;
            fast = fast.next.next;
        }
        
        TreeNode node = new TreeNode(slow.val);
        node.left = bst(head, slow);
        node.right = bst(slow.next, end);
        return node;
    }
}