Valid BST – HackerRank Solution

In this post, we will solve Valid BST HackerRank Solution. This problem (Valid BST) is a part of HackerRank Functional Programming series.

Task

A binary tree is a tree where each node has at most two children. It is characterized by any of the following properties:

1. It can be an empty tree, where root = null.
2. It can contain a root node which contain some value and two subtree, left subtree and right subtree, which are also binary tree.

A binary tree is a binary search tree (BST) if all the non-empty nodes follows both two properties:

1. Each node’s left subtree contains only values less than it, and
2. Each node’s right subtree contains only values greater than it.

Preorder traversal is a tree traversal method where the current node is visited first, then the left subtree and then the right subtree. More specifically, let’s represent the preorder traversal of a tree by a list. Then this list is constructed in following way:

1. If the tree is empty, then this list be a null list.
2. For non-empty tree, let’s represent the preorder of left subtree as L and of right subtree as R. Then the preorder of tree is obtained by appending L to current node, and then appending R to it.

1         2          3
 \       / \        / \
  3     1   3      2   5
 /                /   / \
2                1   4   6
(a)       (b)        (c)

For the above trees, preorder will be

(a) 1 3 2
(b) 2 1 3
(c) 3 2 1 5 4 6

Given a list of numbers, determine whether it can represent the preorder traversal of a binary search tree(BST).

Input

The first line contains the number of test cases, T. Then T test cases follow. The first line of each test case contains the number of nodes in the tree, N. In next line there will a list of N unique numbers, where each number is from set [1, N].

Output

For each test case, print “YES” if there’s exist a BST whose preorder is equal to the list otherwise “NO” (without quotes).

Constraints

• 1 <= T <= 10
• 1 <= N <= 100

Sample Input

5
3
1 2 3
3
2 1 3
6
3 2 1 5 4 6
4
1 3 4 2
5
3 4 5 1 2

Sample Output

YES
YES
YES
NO
NO

Explanation

First three cases are from examples. And last two test cases are invalid because the subtree for 3 is not valid as 2 and 4 are in the wrong order.

Solution – Valid BST – HackerRank Solution

Scala

import java.util.Scanner

trait Tree {
  def value: Int
}

object Tree {
  def parse(values: List[Int]): Boolean = {
    def inner(values: List[Int], lowerBound: Int = Int.MinValue, upperBound: Int = Int.MaxValue): (Tree, List[Int]) = values match {
      case v :: vs if v > lowerBound && v < upperBound =>
        val (left, leftRest) = inner(vs, lowerBound, v)
        val (right, rightRest) = inner(leftRest, v, upperBound)
        (new Node(v, left, right), rightRest)
      case _ => (Empty, values)
    }

    inner(values)._2 == Nil
  }
}

class Node(val value: Int, left: Tree, right: Tree) extends Tree

object Empty extends Tree {
  override def value: Int = throw new Exception("Value of empty tree")
}

object Solution {
  def main(args: Array[String]): Unit = {
    val sc = new Scanner(System.in)

    val t = sc.nextInt
    (0 until t).foreach(_ => {
      val n = sc.nextInt

      val values = (0 until n).map(_ => sc.nextInt).toList

      println(if (Tree.parse(values)) "YES" else "NO")
    })

    sc.close()
  }
}

Note: This problem (Valid BST) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.