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'

Anext

B.next

C->next

D->prev

Attempts:

3 left

💡 Hint

Common Mistakes

Using dot operator instead of arrow operator for pointer access.

Trying to access a member that doesn't exist like 'prev'.

In C, to access the next node in a linked list through a pointer, use the arrow operator -> followed by the member name next.

2fill in blank

medium

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

DSA C

printf("%d ", current[1]);

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A->data

Bdata

C.data

Ddata->

Attempts:

3 left

💡 Hint

Common Mistakes

Using dot operator instead of arrow operator.

Forgetting to access the data member.

Since current is a pointer to a node, use ->data to access the integer data inside the node.

3fill in blank

hard

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

DSA C

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 equality operator == which stops at the first node.

The loop should continue while current is not NULL, so use the not equal operator !=.

4fill in blank

hard

Fill both blanks to complete the for loop that traverses and prints the linked list.

DSA C

for (struct Node* current = head; current [1] NULL; current [2]) {
    printf("%d ", current->data);
}

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

A!=

B==

Ccurrent = current->next

Dcurrent++

Attempts:

3 left

💡 Hint

Common Mistakes

Using current++ which is invalid for pointers to nodes.

Using equality operator == in the loop condition.

The loop continues while current != NULL and moves to the next node with current = current->next.

5fill in blank

hard

Fill all three blanks to create a while loop that traverses and prints the linked list.

DSA C

struct Node* current = [1];
while (current [2] NULL) {
    printf("%d ", current->[3]);
    current = current->next;
}

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

Ahead

B==

Cdata

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

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

Printing a wrong member instead of data.

Start from head, loop while current != NULL, and print current->data.

Practice

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

easy

A. To delete the entire linked list

B. To visit each node and access its data

C. To create a new linked list

D. To reverse the linked list

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

   The main goal is to access or print the data stored in each node.

3. Final Answer:

   To visit each node and access its data -> Option B

4. Quick Check:

   Traversal = Visit nodes [OK]

Hint: Traversal means visiting nodes to read data [OK]

Common Mistakes:

• Confusing traversal with deletion
• Thinking traversal creates new nodes
• Assuming traversal reverses the list

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

easy

A. Node* temp = &head;

B. struct Node temp = head;

C. struct Node* temp = head;

D. Node temp = head;

Solution

1. Step 1: Understand pointer usage

   In C, linked list nodes are accessed via pointers, so temp must be a pointer to Node.

2. Step 2: Check syntax correctness

   struct Node* temp = head; correctly declares temp as a pointer and assigns it to head.

3. Final Answer:

   struct Node* temp = head; -> Option C

4. Quick Check:

   Pointer assignment = struct Node* temp = head [OK]

Hint: Use pointer variable to traverse linked list [OK]

Common Mistakes:

• Using struct Node instead of pointer
• Taking address of head incorrectly
• Missing struct keyword if needed

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

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

int main() {
  struct Node n3 = {30, NULL};
  struct Node n2 = {20, &n3};
  struct Node n1 = {10, &n2};
  struct Node* head = &n1;

  struct Node* temp = head;
  while (temp != NULL) {
    printf("%d -> ", temp->data);
    temp = temp->next;
  }
  printf("NULL\n");
  return 0;
}

medium

A. 10 -> 20 -> 30 -> NULL

B. 30 -> 20 -> 10 -> NULL

C. 10 20 30 NULL

D. Compilation error

Solution

1. Step 1: Trace linked list nodes

   The list nodes are n1(10) -> n2(20) -> n3(30) -> NULL.

2. Step 2: Follow traversal and print

   The loop prints each node's data followed by ' -> ', then prints 'NULL' at the end.

3. Final Answer:

   10 -> 20 -> 30 -> NULL -> Option A

4. Quick Check:

   Print nodes in order = 10 -> 20 -> 30 -> NULL [OK]

Hint: Follow next pointers to print nodes in order [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing '->' arrows in output
• Forgetting to print NULL at end

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

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

void printList(struct Node* head) {
  while (head->next != NULL) {
    printf("%d -> ", head->data);
    head = head->next;
  }
  printf("NULL\n");
}

medium

A. head should be a double pointer

B. Missing semicolon after printf

C. printf format specifier is incorrect

D. The loop condition should be head != NULL

Solution

1. Step 1: Analyze loop condition

   The loop stops when head->next is NULL, so it skips printing the last node.

2. Step 2: Correct loop condition

   Changing condition to head != NULL ensures all nodes, including last, are printed.

3. Final Answer:

   The loop condition should be head != NULL -> Option D

4. Quick Check:

   Loop until head is NULL to print all nodes [OK]

Hint: Loop until pointer is NULL, not next pointer [OK]

Common Mistakes:

• Stopping loop too early
• Using wrong pointer type
• Ignoring last node in print

5. Given a linked list where some nodes may have data value 0, which of the following code snippets correctly prints all nodes including those with 0?

struct Node* temp = head;
while (temp) {
  if (temp->data) {
    printf("%d -> ", temp->data);
  }
  temp = temp->next;
}
printf("NULL\n");

hard

A. Change 'if (temp->data)' to always print without condition

B. Add 'temp = temp->next;' inside the if block only

C. Use 'while (temp != NULL)' and keep the if condition

D. Replace 'printf' with 'puts' to print zeros

Solution

1. Step 1: Understand the if condition effect

   The condition 'if (temp->data)' skips printing nodes with data 0 because 0 is false.

2. Step 2: Fix to print all nodes

   Removing the if condition prints all nodes including those with 0.

3. Final Answer:

   Change 'if (temp->data)' to always print without condition -> Option A

4. Quick Check:

   Print all nodes regardless of data value [OK]

Hint: Do not skip nodes with zero data when printing [OK]

Common Mistakes:

• Skipping zero values due to if condition
• Moving pointer only inside if block causing infinite loop
• Using wrong print function for integers