[LeetCode] 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.

 Thought process:
Divide and conquer. Recursively find the middle of the list and attach left and right sub-tree. Use an instance variable as current pointer to avoid looking for middle point in every recurrence.

 Solution:
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/**
 * 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 {
    private ListNode current;
    
    public TreeNode sortedListToBST(ListNode head) {
        current = head;
        int length = 0;
        
        while (current != null) {
            length++;
            current = current.next;
        }
        
        current = head;
        
        return sortedListToBST(length);
    }
    
    private TreeNode sortedListToBST(int length) {
        if (length == 0) {
            return null;
        }
        
        TreeNode left = sortedListToBST(length / 2);
        TreeNode root = new TreeNode(current.val);
        current = current.next;
        TreeNode right = sortedListToBST(length - length / 2 - 1);
        
        root.left = left;
        root.right = right;
        
        return root;
    }
}

Time complexity: 
Current moves from the start to the end only once. The time complexity is O(n).
Location: San Jose, CA, USA

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