Tree vs Array vs Linked List When Hierarchy Matters in DSA Java - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an array of integers with size 5.

DSA Java

int[] numbers = new int[[1]];

Drag options to blanks, or click blank then click option'

A5

B10

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as size creates an empty array.

Using 1 or 10 creates arrays of wrong size.

The array size must be 5 to hold five integers.

2fill in blank

medium

Complete the code to add a new node with value 10 at the end of a singly linked list.

DSA Java

Node newNode = new Node([1]);
current.next = newNode;

Drag options to blanks, or click blank then click option'

A0

B10

Cnull

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using null instead of a value causes errors.

Using 0 or 5 does not match the required value.

The new node should hold the value 10 as specified.

3fill in blank

hard

Fix the error in the tree node creation by completing the missing child assignment.

DSA Java

TreeNode root = new TreeNode(1);
root.left = new TreeNode([1]);

Drag options to blanks, or click blank then click option'

Anull

Broot

C2

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning null means no child node is created.

Using 0 or root causes logical errors.

The left child node should have value 2 to build the tree correctly.

4fill in blank

hard

Fill both blanks to create a linked list node and link it properly.

DSA Java

Node newNode = new Node([1]);
current.[2] = newNode;

Drag options to blanks, or click blank then click option'

A15

Bnext

Cprev

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' breaks singly linked list logic.

Using 'data' as a pointer is incorrect.

The new node holds value 15 and is linked by setting current.next to newNode.

5fill in blank

hard

Fill all three blanks to create a tree node with value 20 and assign its left and right children.

DSA Java

TreeNode node = new TreeNode([1]);
node.left = new TreeNode([2]);
node.right = new TreeNode([3]);

Drag options to blanks, or click blank then click option'

A20

B15

C25

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using null for children creates missing nodes.

Swapping left and right child values breaks tree order.

The root node has value 20, left child 15, and right child 25 to form a proper binary tree.

Practice

1. Which data structure is best suited to represent hierarchical relationships like a family tree?

easy

A. Tree

B. Array

C. Linked List

D. Stack

Solution

1. Step 1: Understand hierarchy representation

   Hierarchies have parent-child relationships, which trees naturally model with nodes and branches.

2. Step 2: Compare with other structures

   Arrays and linked lists are linear and do not represent branching relationships well.

3. Final Answer:

   Tree -> Option A

4. Quick Check:

   Hierarchy = Tree [OK]

Hint: Hierarchy means branching, so choose Tree [OK]

Common Mistakes:

• Choosing Array because it stores elements
• Choosing Linked List thinking it can represent hierarchy
• Confusing Stack with hierarchical structure

2. Which of the following is the correct way to declare a simple array of integers in Java?

easy

A. array int arr = new int[5];

B. int arr = new int[5];

C. int arr[] = new int();

D. int[] arr = new int[5];

Solution

1. Step 1: Check Java array syntax

   Java arrays are declared with type[], variable name, and size using new keyword.

2. Step 2: Validate each option

   int[] arr = new int[5]; matches correct syntax: int[] arr = new int[5]; others have syntax errors.

3. Final Answer:

   int[] arr = new int[5]; -> Option D

4. Quick Check:

   Java array declaration = int[] arr = new int[5]; [OK]

Hint: Remember Java arrays use type[] name = new type[size]; [OK]

Common Mistakes:

• Missing [] after type
• Using parentheses instead of brackets
• Incorrect keyword 'array'

3. You want to store employee hierarchy where each employee can have multiple subordinates. Which data structure is best and why?

easy

A. Array, because it stores elements in order

B. Linked List, because it links employees one after another

C. Tree, because it models multiple children per node

D. Stack, because it manages last-in-first-out order

Solution

1. Step 1: Understand employee hierarchy needs

   Each employee can have multiple subordinates, so structure must support multiple children per node.

2. Step 2: Evaluate data structures

   Array and linked list are linear, stack is LIFO. Only tree supports multiple children naturally.

3. Final Answer:

   Tree, because it models multiple children per node -> Option C

4. Quick Check:

   Multiple children = Tree [OK]

Hint: Multiple children per node? Choose Tree [OK]

Common Mistakes:

• Choosing Array or Linked List for hierarchical data
• Confusing Stack with hierarchy management
• Ignoring multiple children requirement

4. What will be the output of the following Java code snippet?

class Node {
  int data;
  Node next;
  Node(int d) { data = d; next = null; }
}

public class Test {
  public static void main(String[] args) {
    Node head = new Node(1);
    head.next = new Node(2);
    head.next.next = new Node(3);
    Node current = head;
    while(current != null) {
      System.out.print(current.data + " -> ");
      current = current.next;
    }
    System.out.print("null");
  }
}

medium

A. 1 -> 2 -> 3 -> null

B. 1 2 3 null

C. null -> 1 -> 2 -> 3

D. 1 -> 3 -> 2 -> null

Solution

1. Step 1: Trace linked list creation

   Nodes created with values 1, 2, 3 linked in order: 1 -> 2 -> 3.

2. Step 2: Trace print loop

   Loop prints each node's data followed by ' -> ', ends with 'null'. Output is '1 -> 2 -> 3 -> null'.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option A

4. Quick Check:

   Linked list print = 1 -> 2 -> 3 -> null [OK]

Hint: Follow next pointers to print nodes in order [OK]

Common Mistakes:

• Mixing node order
• Missing 'null' at end
• Printing values without arrows

5. Identify the error in this Java code that tries to create a simple tree node:

class TreeNode {
  int val;
  TreeNode left, right;
  TreeNode(int v) {
    val = v;
    left = null;
    right = null;
  }
}

public class Main {
  public static void main(String[] args) {
    TreeNode root = new TreeNode(10);
    root.left = new TreeNode(5);
    root.right = new TreeNode(15);
    root.left.left = 7;
  }
}

medium

A. Missing semicolon after TreeNode declarations

B. Assigning integer 7 directly to left.left instead of TreeNode

C. Incorrect constructor name in TreeNode class

D. Using 'root' without initialization

Solution

1. Step 1: Analyze assignment to left.left

   left.left expects a TreeNode object but is assigned integer 7 directly, causing type error.

2. Step 2: Check other code parts

   Constructor name and semicolons are correct; root is initialized properly.

3. Final Answer:

   Assigning integer 7 directly to left.left instead of TreeNode -> Option B

4. Quick Check:

   TreeNode fields need TreeNode objects, not int [OK]

Hint: Tree nodes must link to nodes, not raw values [OK]

Common Mistakes:

• Assigning raw values to node references
• Ignoring type mismatch errors
• Assuming automatic conversion from int to TreeNode