Traversal and Printing a Linked List in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

current = current[1];

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

A->next

B.next

Cnext

D->prev

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.' instead of '->' to access next node.

Using 'next' without any operator.

Using '->prev' which points to previous node.

In C++, to access the next node pointer from a node pointer, use '->next'.

2fill in blank

medium

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

DSA C++

std::cout << current[1] << " -> ";

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

Adata

B->data

C.data

D->value

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.' instead of '->' to access data.

Using 'data' without any operator.

Using '->value' which is not the correct member.

Use '->data' to access the data member of the node pointer.

3fill in blank

hard

Fix the error in the loop condition to traverse the linked list correctly.

DSA C++

while (current [1] nullptr) {
    // process node
    current = current->next;
}

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

A==

B=

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' which assigns instead of compares.

Using '==' which stops at first node.

Using '<' which is invalid for pointers.

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

4fill in blank

hard

Fill both blanks to create a loop that prints all node data followed by 'null'.

DSA C++

while (current [1] nullptr) {
    std::cout << current[2] << " -> ";
    current = current->next;
}
std::cout << "null" << std::endl;

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

A!=

B->data

C==

D.data

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' in loop condition causing no iteration.

Using '.' instead of '->' to access data.

Use '!=' to loop until current is nullptr and '->data' to print node data.

5fill in blank

hard

Fill all three blanks to create a function that prints a linked list.

DSA C++

void printList(Node* [1]) {
    Node* current = [2];
    while (current [3] nullptr) {
        std::cout << current->data << " -> ";
        current = current->next;
    }
    std::cout << "null" << std::endl;
}

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

Ahead

C!=

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong parameter name.

Assigning wrong variable to current.

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

The function parameter is 'head', assign it to 'current', and loop while current != nullptr.

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