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.

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

        // new head is slow.val
        TreeNode thead = new TreeNode(slow.val);
        thead.left = binaryTraversal(head, slow);
        thead.right = binaryTraversal(slow.next, tail);
        return thead;
    }
}