Traversal and Printing a Linked List in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to move to the next node in the linked list.

DSA Java

current = current[1];

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

A.next

B.prev

C.head

D.tail

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev instead of .next

Using .head or .tail which are not pointers for traversal

To traverse a linked list, you move from the current node to the next node using the .next pointer.

2fill in blank

medium

Complete the code to print the data of the current node.

DSA Java

System.out.println(current[1]);

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

A.next

B.prev

C.data

D.head

Attempts:

3 left

💡 Hint

Common Mistakes

Printing .next or .prev which are pointers, not data

Trying to print .head which is not part of the node

The data stored in a node is accessed using the .data field.

3fill in blank

hard

Fix the error in the loop condition to traverse until the end of the list.

DSA Java

while (current [1] null) {

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

A==

B=

C=>

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using assignment operator '=' instead of comparison

Using '==' which would stop at the first node

Using invalid operator '=>'

The loop should continue while current is not null, so use !=.

4fill in blank

hard

Fill both blanks to complete the loop that prints all nodes in the linked list.

DSA Java

while (current [1] null) {
    System.out.println(current[2]);
    current = current.next;
}

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

A!=

B.next

C.data

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '!=' in the loop condition

Printing current.next instead of current.data

The loop continues while current != null. Inside the loop, print current.data.

5fill in blank

hard

Fill all three blanks to create a method that prints all elements of a linked list.

DSA Java

public void printList(Node [1]) {
    Node current = [2];
    while (current [3] null) {
        System.out.println(current.data);
        current = current.next;
    }
}

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

Ahead

C!=

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong parameter name

Using '=' instead of '!=' in the loop condition

Assigning wrong variable to current

The method takes the head node as a parameter, assigns it to current, and loops while current is not null.

Practice

1. What is the main purpose of traversing a linked list in Java?

easy

A. To sort the linked list nodes automatically

B. To delete the entire linked list at once

C. To create a new linked list from scratch

D. To visit each node and perform an action like printing its value

Solution

1. Step 1: Understand traversal meaning

   Traversal means visiting each node one by one in the linked list.

2. Step 2: Identify traversal purpose

   Commonly, traversal is used to perform actions like printing or searching nodes.

3. Final Answer:

   To visit each node and perform an action like printing its value -> Option D

4. Quick Check:

   Traversal = Visit nodes [OK]

Hint: Traversal means visiting nodes one by one [OK]

Common Mistakes:

• Confusing traversal with deletion
• Thinking traversal creates a new list
• Assuming traversal sorts the list

2. Which of the following is the correct syntax to start traversing a singly linked list named head in Java?

easy

A. Node current = head.next;

B. Node current = null;

C. Node current = head;

D. Node current = head.prev;

Solution

1. Step 1: Identify starting point of traversal

   Traversal starts from the first node, which is referenced by head.

2. Step 2: Choose correct initialization

   We assign current to head to begin traversal at the first node.

3. Final Answer:

   Node current = head; -> Option C

4. Quick Check:

   Start traversal at head node [OK]

Hint: Start traversal from head node reference [OK]

Common Mistakes:

• Starting from head.next skips first node
• Initializing current to null causes no traversal
• Using head.prev in singly linked list causes error

3. 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 Main {
  public static void main(String[] args) {
    Node head = new Node(5);
    head.next = new Node(10);
    head.next.next = new Node(15);

    Node current = head;
    while (current != null) {
      System.out.print(current.data + " -> ");
      current = current.next;
    }
    System.out.print("null");
  }
}

medium

A. 5 -> 10 -> 15 -> null

B. null -> 5 -> 10 -> 15

C. 5 10 15 null

D. 5 -> 10 -> null

Solution

1. Step 1: Trace node creation and linking

   Nodes created with data 5, 10, 15 linked as 5 -> 10 -> 15 -> null.

2. Step 2: Follow while loop printing

   Loop prints each node's data followed by ' -> ', then ends with 'null'.

3. Final Answer:

   5 -> 10 -> 15 -> null -> Option A

4. Quick Check:

   Print nodes in order with arrows [OK]

Hint: Print each node data with arrow until null [OK]

Common Mistakes:

• Printing null at start instead of end
• Missing arrows between nodes
• Stopping before last node

4. Identify the error in the following code snippet for printing a linked list and select the correct fix.

Node current = head;
while (current.next != null) {
  System.out.print(current.data + " -> ");
  current = current.next;
}
System.out.print("null");

medium

A. Remove System.out.print("null") line

B. Change condition to current != null to print last node

C. Initialize current to head.next to skip first node

D. Add current = current.next inside the loop

Solution

1. Step 1: Analyze loop condition

   The loop stops when current.next is null, so last node is not printed.

2. Step 2: Fix condition to include last node

   Changing condition to current != null ensures last node is printed.

3. Final Answer:

   Change condition to current != null to print last node -> Option B

4. Quick Check:

   Loop until current is null to print all nodes [OK]

Hint: Loop until current is null to include last node [OK]

Common Mistakes:

• Using current.next != null misses last node
• Skipping first node by wrong initialization
• Removing null print causes unclear output

5. Given a singly linked list, how would you modify the traversal code to print the list in reverse order without changing the list structure?

hard

A. Use recursion to traverse to the end, then print nodes on return

B. Reverse the list first, then traverse normally

C. Use a loop to print nodes from tail to head directly

D. Print nodes while moving forward, then print again backward

Solution

1. Step 1: Understand limitation of singly linked list

   Singly linked list nodes only point forward, so backward traversal is not direct.

2. Step 2: Use recursion to print in reverse

   Recursion goes to end first, then prints nodes on returning back, achieving reverse print without changing list.

3. Final Answer:

   Use recursion to traverse to the end, then print nodes on return -> Option A

4. Quick Check:

   Recursion prints reverse without modifying list [OK]

Hint: Recursion prints reverse by returning from end [OK]

Common Mistakes:

• Trying to loop backward in singly linked list
• Reversing list changes structure, not allowed here
• Printing twice forward and backward duplicates output