Traversal and Printing a Linked List in DSA C - Build from Scratch

Traversal and Printing a Linked List

📖 Scenario: Imagine you have a chain of boxes, each holding a number and a link to the next box. This is like a linked list. You want to look inside each box and write down the numbers in order.

🎯 Goal: You will create a simple linked list with three nodes, then write code to go through each node one by one and print the numbers stored inside.

📋 What You'll Learn

Create a struct called Node with an integer data and a pointer next to the next node

Create three nodes with data values 10, 20, and 30 linked together

Create a pointer called head that points to the first node

Write a loop to traverse the linked list starting from head

Print the data of each node followed by ->, ending with NULL

💡 Why This Matters

🌍 Real World

Linked lists are used in many programs to store items in order when the number of items can change. Traversing and printing helps us see the list contents.

💼 Career

Understanding linked lists and how to traverse them is a key skill for software developers, especially when working with low-level data structures or preparing for coding interviews.

1

Create the linked list nodes

Create a struct called Node with an integer data and a pointer next to Node. Then create three variables node1, node2, and node3 of type Node. Set their data values to 10, 20, and 30 respectively.

DSA C

#include 

// Define the Node struct and create node1, node2, node3
// Your code here

Hint

Start by defining the Node struct with two members: data and next. Then create three variables of this struct and assign the data values.

2

Link the nodes and create the head pointer

Set node1.next to point to &node2, node2.next to point to &node3, and node3.next to NULL. Then create a pointer head of type struct Node* and set it to point to &node1.

DSA C

#include 

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

int main() {
    struct Node node1, node2, node3;
    node1.data = 10;
    node2.data = 20;
    node3.data = 30;

    // Link nodes and create head pointer
    // Your code here

    return 0;
}

Hint

Remember to link each node's next to the address of the following node. The last node's next should be NULL. Then set head to the first node's address.

3

Traverse the linked list

Create a pointer current of type struct Node* and set it to head. Use a while loop that runs while current is not NULL. Inside the loop, print the data of current followed by ->. Then update current to current->next.

DSA C

#include 

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

int main() {
    struct Node node1, node2, node3;
    node1.data = 10;
    node2.data = 20;
    node3.data = 30;

    node1.next = &node2;
    node2.next = &node3;
    node3.next = NULL;

    struct Node* head = &node1;

    // Traverse the list and print data
    // Your code here

    return 0;
}

Hint

Use a pointer to move through the list. Print each node's data followed by an arrow. Stop when you reach NULL.

4

Print NULL at the end

After the while loop, print NULL followed by a newline to show the end of the linked list.

DSA C

#include 

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

int main() {
    struct Node node1, node2, node3;
    node1.data = 10;
    node2.data = 20;
    node3.data = 30;

    node1.next = &node2;
    node2.next = &node3;
    node3.next = NULL;

    struct Node* head = &node1;

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

    // Print NULL to end the list
    // Your code here

    return 0;
}

Hint

After the loop finishes, print NULL and a newline to show the list ends.

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