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.

Top-down approach will take n^2 time, cause each time you need to find the root in the linked list, then recursively construct the two parts.

Bottom-Up approach, you need to keep track of from where to build the sub tree and for how many nodes this sub tree has. so the current is used for this purpose.

/**
 * 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; }
 * }
 */
public class Solution {
    ListNode cur = null;
    public TreeNode sortedListToBST(ListNode head) {
        int size = getLen(head);
        cur = head;

        return getBST(size);
    }

    private TreeNode getBST(int size){
        if(size <=0) return null;
        TreeNode left = getBST(size/2);
        TreeNode root = new TreeNode(cur.val);
        cur = cur.next;
        TreeNode right = getBST(size -1 - size/2);
        root.left = left;
        root.right = right;
        return root;
    }
    private int getLen(ListNode head){
        int l = 0;
        while(head != null){
            head = head.next;
            l++;
        }

        return l;
    }
}