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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Traversing a Linked List

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

DSA C++

struct Node {
    int data;
    Node* next;
};

int main() {
    Node n3 = {30, nullptr};
    Node n2 = {20, &n3};
    Node n1 = {10, &n2};

    Node* head = &n1;
    while (head != nullptr) {
        std::cout << head->data << " -> ";
        head = head->next;
    }
    std::cout << "null" << std::endl;
    return 0;
}

A 10 20 30 null

B 30 -> 20 -> 10 -> null

C null -> 10 -> 20 -> 30

D 10 -> 20 -> 30 -> null

💡 Hint

Remember the traversal starts from the head and moves to next nodes until nullptr.

The code starts at node n1 with data 10, then moves to n2 (20), then n3 (30), printing each data followed by ' -> '. Finally, it prints 'null'.

❓ Predict Output

intermediate

Output After Modifying Linked List Pointer

What is the output of this C++ code that modifies the head pointer before traversal?

DSA C++

struct Node {
    int data;
    Node* next;
};

int main() {
    Node n3 = {30, nullptr};
    Node n2 = {20, &n3};
    Node n1 = {10, &n2};

    Node* head = &n2; // start from second node
    while (head != nullptr) {
        std::cout << head->data << " -> ";
        head = head->next;
    }
    std::cout << "null" << std::endl;
    return 0;
}

A 20 -> 30 -> null

B 10 -> 20 -> 30 -> null

C 30 -> null

D null

💡 Hint

The traversal starts from the node with data 20, not 10.

The head pointer is set to n2, so traversal prints 20, then 30, then null.

🔧 Debug

advanced

Identify the Runtime Error in Linked List Traversal

What runtime error will occur when running this C++ code that attempts to traverse a linked list?

DSA C++

struct Node {
    int data;
    Node* next;
};

int main() {
    Node* head = nullptr;
    while (head->next != nullptr) {
        std::cout << head->data << " -> ";
        head = head->next;
    }
    std::cout << "null" << std::endl;
    return 0;
}

A Segmentation fault (accessing nullptr)

B No output, program ends normally

C Compilation error: 'head' undeclared

D Infinite loop

💡 Hint

Check what happens when head is nullptr and you try to access head->next.

The code tries to access head->next when head is nullptr, causing a segmentation fault (runtime crash).

❓ Predict Output

advanced

Output of Recursive Linked List Print

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

DSA C++

struct Node {
    int data;
    Node* next;
};

void printList(Node* node) {
    if (node == nullptr) {
        std::cout << "null" << std::endl;
        return;
    }
    std::cout << node->data << " -> ";
    printList(node->next);
}

int main() {
    Node n3 = {30, nullptr};
    Node n2 = {20, &n3};
    Node n1 = {10, &n2};
    printList(&n1);
    return 0;
}

A null -> 10 -> 20 -> 30

B 10 -> 20 -> 30 -> null

C 10 20 30 null

D 30 -> 20 -> 10 -> null

💡 Hint

The function prints current node data then calls itself on next node.

The recursion prints nodes in order: 10, 20, 30, then prints 'null' at the end.

🧠 Conceptual

expert

Number of Nodes in a Linked List After Traversal

Given a singly linked list with nodes containing values 5, 15, 25, 35 linked in order, what is the number of nodes counted after traversing the list from head to end?

A 3

B 0

C 4

D 5

💡 Hint

Count each node visited until you reach nullptr.

The linked list has nodes with values 5, 15, 25, 35 linked in order, which is 4 nodes.

Practice

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

easy

A. To delete all nodes at once

B. To reverse the linked list automatically

C. To create a new linked list

D. To visit each node and access its data

Solution

1. Step 1: Understand traversal meaning

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

2. Step 2: Purpose of traversal

   The main goal is to access or process the data stored in each node during the visit.

3. Final Answer:

   To visit each node and access its data -> 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 new lists
• Assuming traversal reverses the list

2. Which of the following is the correct syntax to start traversing a singly linked list in C++ if head is the pointer to the first node?

easy

A. Node* current = head;

B. Node current = head;

C. Node* current = &head;

D. Node current = *head;

Solution

1. Step 1: Understand pointer usage

   In C++, linked list nodes are accessed via pointers. head is a pointer to the first node.

2. Step 2: Correct pointer assignment

   To traverse, we create a pointer current and assign it to head directly: Node* current = head;

3. Final Answer:

   Node* current = head; -> Option A

4. Quick Check:

   Use pointer to start traversal [OK]

Hint: Use pointer assignment to start traversal [OK]

Common Mistakes:

• Using object instead of pointer
• Taking address of head incorrectly
• Dereferencing head wrongly

3. What will be the output of the following C++ code snippet that prints a linked list?

struct Node {
  int data;
  Node* next;
};

int main() {
  Node n3 = {3, nullptr};
  Node n2 = {2, &n3};
  Node n1 = {1, &n2};
  Node* head = &n1;

  Node* current = head;
  while (current != nullptr) {
    std::cout << current->data << " -> ";
    current = current->next;
  }
  std::cout << "null" << std::endl;
  return 0;
}

medium

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

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

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

D. 1 2 3 null

Solution

1. Step 1: Trace node connections

   Node n1 points to n2, n2 points to n3, n3 points to nullptr (end).

2. Step 2: Follow the while loop output

   The loop prints data of current node then moves to next. So prints 1, then 2, then 3, then stops.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option C

4. Quick Check:

   Print nodes in order from head [OK]

Hint: Follow next pointers from head until nullptr [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing '->' arrows in output
• Not printing 'null' at the end

4. Identify the error in the following code snippet that attempts to print a linked list:

struct Node {
  int data;
  Node* next;
};

void printList(Node* head) {
  Node* current = head;
  while (current->next != nullptr) {
    std::cout << current->data << " -> ";
    current = current->next;
  }
  std::cout << "null" << std::endl;
}

medium

A. The loop condition should be current != nullptr, not current->next != nullptr

B. The print statement should be after the loop

C. The function should return an int

D. The pointer current should be declared as Node current, not Node* current

Solution

1. Step 1: Analyze loop condition

   The loop stops when current->next is nullptr, so it skips printing the last node's data.

2. Step 2: Correct loop condition

   It should check current != nullptr to print all nodes including the last one.

3. Final Answer:

   The loop condition should be current != nullptr, not current->next != nullptr -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Stopping loop too early missing last node
• Wrong pointer type declaration
• Misplacing print statement

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 or using recursion?

hard

A. Traverse normally and print nodes as you go

B. Use a stack to store nodes during traversal, then pop and print

C. Reverse the list first, then print

D. Use two pointers to print backwards

Solution

1. Step 1: Understand constraints

   We cannot change the list or use recursion, so reversing pointers or recursive calls are not allowed.

2. Step 2: Use auxiliary data structure

   Use a stack to push each node's data during traversal, then pop from stack to print in reverse order.

3. Final Answer:

   Use a stack to store nodes during traversal, then pop and print -> Option B

4. Quick Check:

   Stack reverses order without modifying list [OK]

Hint: Use stack to reverse print without changing list [OK]

Common Mistakes:

• Trying to reverse list pointers directly
• Using recursion despite restriction
• Assuming two pointers can print backwards