Traversal and Printing a Linked List in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Traversing a Linked List

What is the output of the following Java code that traverses and prints a singly linked list?

DSA Java

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

public class Main {
    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");
    }
}

A 1 -> 2 -> null

B 1 -> 2 -> 3 -> null

C 1 -> 3 -> null

D null

💡 Hint

Follow the linked list nodes from head until you reach null.

The code creates a linked list with nodes containing 1, 2, and 3. It prints each node's data followed by ' -> ', ending with 'null'.

❓ Predict Output

intermediate

Output When Skipping a Node in Traversal

What is the output of this Java code that traverses a linked list but skips printing the second node?

DSA Java

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

public class Main {
    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) {
            if (current.data == 2) {
                current = current.next;
                continue;
            }
            System.out.print(current.data + " -> ");
            current = current.next;
        }
        System.out.print("null");
    }
}

A 1 -> null

B 1 -> 2 -> 3 -> null

C 2 -> 3 -> null

D 1 -> 3 -> null

💡 Hint

Notice the continue statement skips printing when data is 2.

The code skips printing the node with data 2, so only nodes 1 and 3 are printed followed by 'null'.

🔧 Debug

advanced

Identify the Error in Linked List Traversal

What error will this Java code produce when trying to traverse and print a linked list?

DSA Java

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

public class Main {
    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.next != null) {
            System.out.print(current.data + " -> ");
            current = current.next;
        }
        System.out.print("null");
    }
}

A Output: 1 -> 2 -> null

B Output: 1 -> 2 -> 3 -> null

C NullPointerException at runtime

D Compilation error

💡 Hint

Check the loop condition carefully; it stops before the last node.

The loop condition 'current.next != null' stops before printing the last node (3), so only 1 and 2 are printed before 'null'.

❓ Predict Output

advanced

Output of Recursive Linked List Print

What is the output of this recursive Java method that prints a linked list?

DSA Java

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

public class Main {
    public static void printList(Node node) {
        if (node == null) {
            System.out.print("null");
            return;
        }
        System.out.print(node.data + " -> ");
        printList(node.next);
    }

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

A 5 -> 10 -> 15 -> null

B null

C 5 -> 10 -> null

D 15 -> 10 -> 5 -> null

💡 Hint

The method prints current node then calls itself on next node.

The recursive method prints each node's data followed by ' -> ', then prints 'null' at the end.

🧠 Conceptual

expert

Number of Nodes Printed After Partial Traversal

Given a linked list with nodes containing values 7, 14, 21, 28, 35, what is the number of nodes printed by this Java code snippet?

DSA Java

Node head = new Node(7);
head.next = new Node(14);
head.next.next = new Node(21);
head.next.next.next = new Node(28);
head.next.next.next.next = new Node(35);

Node current = head.next.next; // start at node with 21
int count = 0;
while (current != null) {
    System.out.print(current.data + " -> ");
    current = current.next;
    count++;
}
System.out.print("null");

// What is the value of count after this loop?

A 4

B 5

C 3

D 2

💡 Hint

Count nodes starting from the third node (value 21) to the end.

Starting at node with value 21, the loop prints nodes 21, 28, and 35, so count is 3.

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