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

#include <stdio.h>
#include <stdlib.h>

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

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

int main() {
    struct Node* head = malloc(sizeof(struct Node));
    struct Node* second = malloc(sizeof(struct Node));
    struct Node* third = malloc(sizeof(struct Node));

    head->data = 10;
    head->next = second;

    second->data = 20;
    second->next = third;

    third->data = 30;
    third->next = NULL;

    printList(head);
    return 0;
}

A NULL

B 10 -> 20 -> NULL

C 20 -> 30 -> NULL

D 10 -> 20 -> 30 -> NULL

💡 Hint

Follow the next pointers from head until NULL is reached.

The code creates three nodes linked in order: 10, 20, 30. The printList function prints each node's data followed by '->' until it reaches NULL, then prints 'NULL'.

❓ Predict Output

intermediate

Output After Modifying Linked List Pointer

What will be the output of the following code that modifies the linked list before printing?

DSA C

#include <stdio.h>
#include <stdlib.h>

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

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

int main() {
    struct Node* head = malloc(sizeof(struct Node));
    struct Node* second = malloc(sizeof(struct Node));
    struct Node* third = malloc(sizeof(struct Node));

    head->data = 1;
    head->next = second;

    second->data = 2;
    second->next = third;

    third->data = 3;
    third->next = NULL;

    // Modify the list
    second->next = NULL;

    printList(head);
    return 0;
}

A 1 -> 2 -> 3 -> NULL

B 1 -> 2 -> NULL

C 2 -> 3 -> NULL

D 1 -> 3 -> NULL

💡 Hint

Check how the next pointer of the second node is changed.

The second node's next pointer is set to NULL, so the list ends after 2. The third node is disconnected and not printed.

🔧 Debug

advanced

Identify the Error in Linked List Traversal

What error will occur when running this code that attempts to print a linked list?

DSA C

#include <stdio.h>
#include <stdlib.h>

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

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

int main() {
    struct Node* head = malloc(sizeof(struct Node));
    head->data = 5;
    head->next = NULL;

    printList(head);
    return 0;
}

A No error, output: 5 -> NULL

B Segmentation fault (accessing NULL pointer)

C Infinite loop

D Compilation error

💡 Hint

Check the loop condition and how it handles single-node list.

The loop runs while current->next != NULL. For a single node with next NULL, loop skips and prints the last node's data correctly.

❓ Predict Output

advanced

Output of Recursive Linked List Print

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

DSA C

#include <stdio.h>
#include <stdlib.h>

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

void printListRecursive(struct Node* node) {
    if (node == NULL) {
        printf("NULL\n");
        return;
    }
    printf("%d -> ", node->data);
    printListRecursive(node->next);
}

int main() {
    struct Node* head = malloc(sizeof(struct Node));
    struct Node* second = malloc(sizeof(struct Node));

    head->data = 7;
    head->next = second;

    second->data = 14;
    second->next = NULL;

    printListRecursive(head);
    return 0;
}

A 14 -> 7 -> NULL

B 7 -> 14 ->

C 7 -> 14 -> NULL

D NULL

💡 Hint

Think about how recursion prints nodes in order.

The function prints current node data then calls itself on next node, so output is in list order ending with NULL.

🧠 Conceptual

expert

Number of Nodes After Partial Traversal

Given a linked list with 5 nodes, if you traverse the list with the following code snippet, how many nodes will be counted?

DSA C

int countNodes(struct Node* head) {
    int count = 0;
    struct Node* current = head;
    while (current != NULL && count < 3) {
        count++;
        current = current->next;
    }
    return count;
}

A 3

B 5

C 0

D 4

💡 Hint

The loop stops when count reaches 3 or list ends.

The loop counts nodes but stops after counting 3 nodes even if list has 5 nodes.

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