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.