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 ofeverynode 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

Solution)

By binary search from the list, we can make each left and right node based on the middle point.

space complexity (O(n))

/**
 * 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 helper(ArrayList<ListNode> arr, int start, int end) {
        if (start > end) return null;
        if (start == end) {
            return new TreeNode(arr.get(start).val);
        }
        int m = (start+end)/2;
        TreeNode node = new TreeNode(arr.get(m).val);
        node.left = helper(arr, start, m-1);
        node.right = helper(arr, m+1, end);
        return node;
    }
    public TreeNode sortedListToBST(ListNode head) {
        ArrayList<ListNode> arr = new ArrayList<>();
        ListNode cur = head;
        while(cur != null) {
            arr.add(cur);
            cur = cur.next;
        }
        return helper(arr, 0, arr.size()-1);
    }
}

Solution)

No additional space

class Solution {
    public TreeNode sortedListToBST(ListNode head) {
        if(head==null) return null;
        return toBST(head,null);
    }
    public TreeNode toBST(ListNode head, ListNode tail){
        ListNode slow = head;
        ListNode fast = head;
        if(head==tail) return null;

        while(fast!=tail&&fast.next!=tail){
            fast = fast.next.next;
            slow = slow.next;
        }
        TreeNode thead = new TreeNode(slow.val);
        thead.left = toBST(head,slow);
        thead.right = toBST(slow.next,tail);
        return thead;
    }
}