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

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this linked list traversal code?

Consider a singly linked list with nodes containing values 1, 2, 3 in that order. What will be printed after running the following traversal code?

DSA C#

class Node {
    public int Value;
    public Node Next;
    public Node(int val) { Value = val; Next = null; }
}

Node head = new Node(1);
head.Next = new Node(2);
head.Next.Next = new Node(3);

Node current = head;
while (current != null) {
    Console.Write(current.Value + " -> ");
    current = current.Next;
}
Console.WriteLine("null");

A 1 -> 2 -> 3 -> null

B 3 -> 2 -> 1 -> null

C 1 -> 2 -> null

D null

💡 Hint

Think about how the traversal moves from the head node to the next until it reaches the end.

The code starts at the head node with value 1 and moves through each node using the Next pointer, printing each value followed by ' -> '. It ends when current is null, printing 'null'. So the output is '1 -> 2 -> 3 -> null'.

❓ Predict Output

intermediate

What is the output after modifying the traversal to skip one node?

Given the same linked list as before (1 -> 2 -> 3 -> null), what will be printed after running this modified traversal code?

DSA C#

Node current = head;
while (current != null) {
    Console.Write(current.Value + " -> ");
    if (current.Next != null) {
        current = current.Next.Next;
    } else {
        current = null;
    }
}
Console.WriteLine("null");

A 2 -> 3 -> null

B 1 -> 2 -> 3 -> null

C 1 -> 3 -> null

D 1 -> null

💡 Hint

Notice how the current pointer moves two steps ahead each time if possible.

The traversal prints the current node's value, then moves two nodes ahead if possible. Starting at 1, it prints 1, then moves to node 3 (skipping 2), prints 3, then moves beyond the list (null), ending the loop.

🔧 Debug

advanced

What error does this linked list traversal code cause?

Examine the following code that attempts to print all node values in a linked list. What error will it cause when run?

DSA C#

Node current = head;
while (current.Next != null) {
    Console.Write(current.Value + " -> ");
    current = current.Next;
}
Console.WriteLine("null");

A It prints all nodes except the last one, then prints 'null'.

B NullReferenceException at runtime.

C Compilation error due to missing semicolon.

D Infinite loop.

💡 Hint

Look carefully at the loop condition and what happens to the last node.

The loop condition checks current.Next != null, so it stops before the last node. The last node's value is never printed, but 'null' is printed after the loop. No error occurs.

🧠 Conceptual

advanced

How many nodes are printed by this traversal code?

Given a linked list with 5 nodes, what is the number of nodes printed by this traversal code snippet?

DSA C#

Node current = head;
int count = 0;
while (current != null && count < 3) {
    Console.Write(current.Value + " -> ");
    current = current.Next;
    count++;
}
Console.WriteLine("null");

A 2 nodes

B 5 nodes

C 4 nodes

D 3 nodes

💡 Hint

The loop stops when count reaches 3 or the list ends.

The loop prints nodes while count is less than 3, so it prints exactly 3 nodes even if the list has more nodes.

🚀 Application

expert

What is the output after reversing and printing the linked list?

Given a linked list 1 -> 2 -> 3 -> null, the following code reverses it and then prints the nodes. What is the printed output?

DSA C#

Node prev = null;
Node current = head;
while (current != null) {
    Node nextTemp = current.Next;
    current.Next = prev;
    prev = current;
    current = nextTemp;
}

Node newHead = prev;
while (newHead != null) {
    Console.Write(newHead.Value + " -> ");
    newHead = newHead.Next;
}
Console.WriteLine("null");

A 1 -> 2 -> 3 -> null

B 3 -> 2 -> 1 -> null

C null

D 1 -> null

💡 Hint

The code reverses the links between nodes before printing.

The code reverses the linked list by changing each node's Next pointer to the previous node. After reversal, the list is 3 -> 2 -> 1 -> null, which is printed.

Practice

1. What does traversing a linked list mean in simple terms?

easy

A. Reversing the order of nodes

B. Visiting each node one by one from start to end

C. Deleting the first node

D. Adding a new node at the end

Solution

1. Step 1: Understand traversal

   Traversal means going through each node in the linked list starting from the first node.

2. Step 2: Identify what happens during traversal

   We visit each node one by one until we reach the end (null).

3. Final Answer:

   Visiting each node one by one from start to end -> Option B

4. Quick Check:

   Traversal = Visit nodes sequentially [OK]

Hint: Traversal means visiting nodes one after another [OK]

Common Mistakes:

• Confusing traversal with adding or deleting nodes
• Thinking traversal changes the list order
• Assuming traversal skips nodes

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

easy

A. Node current => head;

B. Node current == head;

C. Node current = head;

D. Node current <- head;

Solution

1. Step 1: Check variable assignment syntax in C#

   In C#, to assign a value, use a single equals sign (=), not double (==) or arrows.

2. Step 2: Identify correct assignment for traversal start

   We assign the head node to a variable to start traversal: Node current = head;

3. Final Answer:

   Node current = head; -> Option C

4. Quick Check:

   Assignment uses =, not == or arrows [OK]

Hint: Use single = for assignment in C# [OK]

Common Mistakes:

• Using == instead of = for assignment
• Using arrow symbols which are invalid
• Confusing assignment with comparison

3. What will be the output of the following C# code snippet?

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

Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);

Node current = head;
while (current != null) {
  Console.Write(current.data + " -> ");
  current = current.next;
}
Console.WriteLine("null");

medium

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

B. null -> 1 -> 2 -> 3

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

D. 1 2 3 null

Solution

1. Step 1: Understand the linked list structure

   The list has nodes with data 1, 2, and 3 linked in order.

2. Step 2: Trace the while loop output

   The loop prints each node's data followed by ' -> ', then moves to next node until null.

3. Final Answer:

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

4. Quick Check:

   Print nodes in order with arrows and end with null [OK]

Hint: Print nodes in order, end with null [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing arrows or null in output
• Printing null at start instead of end

4. Identify the error in this linked list traversal code snippet:

Node current = head;
while (current.next != null) {
  Console.Write(current.data + " -> ");
  current = current.next;
}
Console.WriteLine("null");

medium

A. It prints nodes in reverse order

B. It causes a null reference exception

C. It never enters the loop

D. It skips printing the last node's data

Solution

1. Step 1: Analyze the loop condition

   The loop runs while current.next is not null, so it stops before the last node.

2. Step 2: Check what gets printed

   Since the last node's data is printed only inside the loop, it is skipped.

3. Final Answer:

   It skips printing the last node's data -> Option D

4. Quick Check:

   Loop condition excludes last node [OK]

Hint: Use current != null to include last node [OK]

Common Mistakes:

• Using current.next != null instead of current != null
• Assuming last node prints automatically
• Confusing loop condition with node data access

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 print nodes after recursive call returns

B. Traverse normally and print nodes immediately

C. Reverse the list first then print

D. Use a for loop with index to print backwards

Solution

1. Step 1: Understand the problem constraints

   We cannot change the list structure, so reversing the list is not allowed.

2. Step 2: Use recursion to print in reverse

   Recursion visits nodes forward, then prints after returning, effectively printing backwards.

3. Final Answer:

   Use recursion to print nodes after recursive call returns -> Option A

4. Quick Check:

   Recursion prints in reverse without modifying list [OK]

Hint: Recursion prints reverse by printing after calls return [OK]

Common Mistakes:

• Trying to use index-based loops on linked list
• Reversing list which changes structure
• Printing nodes immediately during traversal