109. Convert Sorted List to Binary Search Tree - Medium

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)　　-- O(n) for recursive call, inside each recursive call, O(logn) for finding middle, 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 toBST(head, null);
    }
    
    private TreeNode toBST(ListNode head, ListNode tail) {
        if(head == tail) return null;
        
        ListNode fast = head, slow = head;
        while(fast.next != tail && fast.next.next != tail) {
            fast = fast.next.next;
            slow = slow.next;
        }
        
        TreeNode root = new TreeNode(slow.val);
        root.left = toBST(head, slow);
        root.right = toBST(slow.next, tail);
        return root;
    }
}