Create and Initialize Doubly Linked List in DSA C - Build from Scratch

Create and Initialize Doubly Linked List

📖 Scenario: You are building a simple program to manage a list of tasks. Each task has a number and you want to store them in a doubly linked list so you can move forward and backward easily.

🎯 Goal: Create a doubly linked list with three nodes containing the values 10, 20, and 30. Initialize the list so that each node points correctly to the previous and next nodes.

📋 What You'll Learn

Define a struct called Node with an integer data and two pointers: prev and next

Create three nodes dynamically with values 10, 20, and 30

Link the nodes so that prev and next pointers are set correctly

Create a pointer called head that points to the first node

💡 Why This Matters

🌍 Real World

Doubly linked lists are used in real-world applications like browser history navigation, music playlists, and undo-redo features where moving forward and backward is needed.

💼 Career

Understanding doubly linked lists is important for software developers working on systems that require efficient bidirectional data traversal and dynamic data management.

1

Define the Node structure and create three nodes

Define a struct called Node with an integer data and two pointers prev and next. Then create three nodes called node1, node2, and node3 using malloc. Assign the values 10, 20, and 30 to node1->data, node2->data, and node3->data respectively.

DSA C

#include 
#include 

int main() {

    // Define struct Node and create three nodes with values 10, 20, 30
    // Your code here

    return 0;
}

Hint

Start by defining the Node struct with data, prev, and next. Then use malloc to create three nodes and assign their data values.

2

Create a head pointer and initialize it

Create a pointer called head of type Node* and set it to point to node1.

DSA C

#include 
#include 

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

int main() {
    Node* node1 = (Node*)malloc(sizeof(Node));
    Node* node2 = (Node*)malloc(sizeof(Node));
    Node* node3 = (Node*)malloc(sizeof(Node));

    node1->data = 10;
    node2->data = 20;
    node3->data = 30;

    // Create head pointer and set it to node1
    // Your code here

    return 0;
}

Hint

Declare head as a pointer to Node and assign it to node1.

3

Link the nodes with prev and next pointers

Set node1->prev to NULL, node1->next to node2. Set node2->prev to node1, node2->next to node3. Set node3->prev to node2, and node3->next to NULL.

DSA C

#include 
#include 

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

int main() {
    Node* node1 = (Node*)malloc(sizeof(Node));
    Node* node2 = (Node*)malloc(sizeof(Node));
    Node* node3 = (Node*)malloc(sizeof(Node));

    node1->data = 10;
    node2->data = 20;
    node3->data = 30;

    Node* head = node1;

    // Link nodes with prev and next pointers
    // Your code here

    return 0;
}

Hint

Remember the first node's prev is NULL and the last node's next is NULL. Link the middle node's pointers to the nodes before and after it.

4

Print the doubly linked list forward

Use a Node* pointer called current to traverse the list from head. Print each node's data followed by ->. After the last node, print NULL.

DSA C

#include 
#include 

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

int main() {
    Node* node1 = (Node*)malloc(sizeof(Node));
    Node* node2 = (Node*)malloc(sizeof(Node));
    Node* node3 = (Node*)malloc(sizeof(Node));

    node1->data = 10;
    node2->data = 20;
    node3->data = 30;

    Node* head = node1;

    node1->prev = NULL;
    node1->next = node2;

    node2->prev = node1;
    node2->next = node3;

    node3->prev = node2;
    node3->next = NULL;

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

    return 0;
}

Hint

Use a while loop to move current from head to the end, printing each data value followed by ->. After the loop, print NULL.