# HackerRank Solutions – Trees – Is This a Binary Search Tree? – Java Solution

All credits to Rodney Shaghoulian for this simple solution for the HackerRank challenge – Trees – Is This a Binary Search Tree. This solution is written in Java.

```// Author: Rodney Shaghoulian
// Github: github.com/RodneyShag

import java.util.Scanner;
import java.util.Stack;

public class Solution {
public static void main(String[] args) {
MyQueue<Integer> queue = new MyQueue<Integer>();

Scanner scan = new Scanner(System.in);
int n = scan.nextInt();

for (int i = 0; i < n; i++) {
int operation = scan.nextInt();
if (operation == 1) { // enqueue
queue.enqueue(scan.nextInt());
} else if (operation == 2) { // dequeue
queue.dequeue();
} else if (operation == 3) { // print/peek
System.out.println(queue.peek());
}
}
scan.close();
}

public static class MyQueue<Integer> {
private Stack<Integer> stack1 = new Stack<>();
private Stack<Integer> stack2 = new Stack<>();

public void enqueue(Integer num) {
stack1.push(num);
}

public Integer dequeue() {
if (size() == 0) {
return null;
}
if (stack2.isEmpty()) {
shiftStacks();
}
return stack2.pop();
}

public Integer peek() {
if (size() == 0) {
return null;
}
if (stack2.isEmpty()) {
shiftStacks();
}
return stack2.peek();
}

/* Only shifts stacks if necessary */
private void shiftStacks() {
if (stack2.isEmpty()) { // shifting items while stack2 contains items would mess up our queue's ordering
while ( ! stack1.isEmpty()) {
stack2.push(stack1.pop());
}
}
}

public int size() {
return stack1.size() + stack2.size();
}
}
}```
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