Print Linked List

Try it on GfG Practice

Given a Singly Linked List, the task is to print all the elements in the list.

Examples:

Input: 1->2->3->4->5->null
Output: 1 2 3 4 5
Explanation: Every element of each node from head node to last node is printed.

Input: 10->20->30->40->50->null
Output: 10 20 30 40 50
Explanation: Every element of each node from head node to last node is printed.

Iterative Approach - O(n) Time and O(1) Space

The process of printing a singly linked list involves printing the value of each node and then going on to the next node and print that node's value also and so on, till we reach the last node in the singly linked list, whose next node points towards the null.

Step-by-Step Algorithm

• We will initialize a temporary pointer to the head node of the singly linked list.
• After that, we will check if that pointer is null or not null, if it is null, then return.
• While the pointer is not null, we will access and print the data of the current node, then we move the pointer to next node.

Program to Print the Singly Linked List using Iteration.

 {...} //Driver Code Starts{
#include <iostream>

using namespace std;

//Driver Code Ends }

// A linked list node
class Node {
public:
    int data;
    Node* next;

    // Constructor to initialize a new node with data
    Node(int new_data) {
        this->data = new_data;
        this->next = nullptr;
    }
};

// Function to print the singly linked list
void printList(Node* head) {

    // A loop that runs till head is nullptr
    while (head != nullptr) {

        // Printing data of current node
        cout << head->data << " ";

        // Moving to the next node
        head = head->next;
    }
}

 {...} //Driver Code Starts{

int main() {
  
    // Create a linked list: 10 -> 20 -> 30 -> 40
    Node* head = new Node(10);
    head->next = new Node(20);
    head->next->next = new Node(30);
    head->next->next->next = new Node(40);

    printList(head);

    return 0;
}

//Driver Code Ends }

 {...} //Driver Code Starts{
#include <stdio.h>
#include <stdlib.h>

//Driver Code Ends }

// A linked list node
struct Node {
    int data;
    struct Node* next;
};

// Function to create a new node
struct Node* createNode(int new_data) {
    struct Node* node = (struct Node*)malloc(sizeof(struct Node));
    node->data = new_data;
    node->next = NULL;
    return node;
}

// Function to print the singly linked list
void printList(struct Node* head) {

    // A loop that runs till head is NULL
    while (head != NULL) {

        // Printing data of current node
        printf("%d ", head->data);

        // Moving to the next node
        head = head->next;
    }
}

 {...} //Driver Code Starts{

int main() {

    // Create a linked list: 10 -> 20 -> 30 -> 40
    struct Node* head = createNode(10);
    head->next = createNode(20);
    head->next->next = createNode(30);
    head->next->next->next = createNode(40);

    printList(head);

    return 0;
}

//Driver Code Ends }

class Node {
    int data;
    Node next;

    // Constructor to initialize a new node with data
    Node(int newData) {
        this.data = newData;
        this.next = null;
    }
}

class GfG {

    // Function to print the singly linked list
    static void printList(Node head) {

        // A loop that runs till head is null
        while (head != null) {

            // Printing data of current node
            System.out.print(head.data + " ");

            // Moving to the next node
            head = head.next;
        }
    }

 {...} 
//Driver Code Starts{
    public static void main(String[] args) {

        // Create a linked list: 10 -> 20 -> 30 -> 40
        Node head = new Node(10);
        head.next = new Node(20);
        head.next.next = new Node(30);
        head.next.next.next = new Node(40);

        printList(head);
    }
}

//Driver Code Ends }

# A linked list node
class Node:
    def __init__(self, newData):
        # Constructor to initialize a new node with data
        self.data = newData
        self.next = None

# Function to print the singly linked list
def printList(head):

    # A loop that runs till head is None
    while head is not None:

        # Printing data of current node
        print(head.data, end=" ")

        # Moving to the next node
        head = head.next

if __name__ == "__main__":

 {...} 
#Driver Code Starts{
    # Create a linked list: 10 -> 20 -> 30 -> 40
    head = Node(10)
    head.next = Node(20)
    head.next.next = Node(30)
    head.next.next.next = Node(40)

    printList(head)

#Driver Code Ends }

 {...} //Driver Code Starts{
using System;

//Driver Code Ends }

class Node {
    public int data;
    public Node next;

    // Constructor to initialize a new node with data
    public Node(int new_data) {
        this.data = new_data;
        this.next = null;
    }
}

class GfG {

    // Function to print the singly linked list
    static void printList(Node head) {

        // A loop that runs till head is null
        while (head != null) {

            // Printing data of current node
            Console.Write(head.data + " ");

            // Moving to the next node
            head = head.next;
        }
    }

 {...} //Driver Code Starts{

    static void Main(string[] args) {

        // Create a linked list: 10 -> 20 -> 30 -> 40
        Node head = new Node(10);
        head.next = new Node(20);
        head.next.next = new Node(30);
        head.next.next.next = new Node(40);

        printList(head);
    }
}

//Driver Code Ends }

class Node {
    constructor(newData) {
        // Constructor to initialize a new node with data
        this.data = newData;
        this.next = null;
    }
}

// Function to print the singly linked list
function printList(head) {

	let result = '';

    // A loop that runs till head is null
    while (head !== null) {

        // Printing data of current node
        result += head.data + ' ';

        // Moving to the next node
        head = head.next;
    }
    
    console.log(result.trim());
}

 {...} 
//Driver Code Starts{
// Driver Code

// Create a linked list: 10 -> 20 -> 30 -> 40
const head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);

printList(head);


//Driver Code Ends }

10 20 30 40 

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(1)

Recursive Approach - O(n) Time and O(n) Memory Space

We can also traverse the singly linked list using recursion. We start at the head node of the singly linked list, check if it is null or not and print its value. We then call the traversal function again with the next node passed as pointer.

Step-by-Step Algorithm

• Firstly, we define a recursive method to traverse the singly linked list, which takes a node as a parameter.
• In this function, the base case is that if the node is null then we will return from the recursive method.
• We then pass the head node as the parameter to this function.
• After that, we access and print the data of the current node.
• At last, we will make a recursive call to this function with the next node as the parameter.

Program to Print the Singly Linked List using Recursion.

#include <iostream>
using namespace std;

// A linked list node
class Node {
public:
    int data;
    Node* next;

    // Constructor to initialize a new node with data
    Node(int new_data) {
        this->data = new_data;
        this->next = nullptr;
    }
};

// Function to print the singly linked list
void printList(Node* head) {

    // Base condition is when the head is nullptr
    if (head == nullptr) {
        return;
    }

    // Printing the current node data
    cout << head->data << " ";

    // Moving to the next node
    printList(head->next);
}

int main() {
  
    // Create a linked list: 10 -> 20 -> 30 -> 40
    Node* head = new Node(10);
    head->next = new Node(20);
    head->next->next = new Node(30);
    head->next->next->next = new Node(40);

    printList(head);

    return 0;
}

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

// A linked list node
struct Node {
    int data;
    struct Node* next;
};

// Function to create a new node with given data
struct Node* createNode(int new_data) {
    struct Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));
    new_node->data = new_data;
    new_node->next = NULL;
    return new_node;
}

// Function to print the singly linked list
void printList(struct Node* head) {
  
    // Base condition is when the head is nullptr
    if (head == NULL) {
        return;
    }

    // Printing the current node data
    printf("%d ", head->data);

    // Moving to the next node
    printList(head->next);
}

int main() {
  
    // Create a linked list: 10 -> 20 -> 30 -> 40
    struct Node* head = createNode(10);
    head->next = createNode(20);
    head->next->next = createNode(30);
    head->next->next->next = createNode(40);

    printList(head);

    return 0;
}

// A linked list node
class Node {
    int data;
    Node next;

    // Constructor to initialize a new node with data
    Node(int new_data) {
        data = new_data;
        next = null;
    }
}

class GfG {

    // Function to print the singly linked list
    static void printList(Node head) {

        // Base condition is when the head is nullptr
        if (head == null) {
            return;
        }

        // Printing the current node data
        System.out.print(head.data + " ");

        // Moving to the next node
        printList(head.next);
    }

    public static void main(String[] args) {
      
        // Create a linked list: 10 -> 20 -> 30 -> 40
        Node head = new Node(10);
        head.next = new Node(20);
        head.next.next = new Node(30);
        head.next.next.next = new Node(40);

        printList(head);
    }
}

# A linked list node
class Node:
    def __init__(self, data):
      
        # Constructor to initialize a new node with data
        self.data = data
        self.next = None

# Function to print the singly linked list
def printList(head):
  
    # Base condition is when the head is nullptr
    if head is None:
        return
      
    # Printing the current node data
    print(head.data, end=" ")
    
    # Moving to the next node
    printList(head.next)

if __name__ == "__main__":
  
    # Create a linked list: 10 -> 20 -> 30 -> 40
    head = Node(10)
    head.next = Node(20)
    head.next.next = Node(30)
    head.next.next.next = Node(40)

    printList(head)

using System;

// A linked list node
class Node {
    public int Data { get;set; }
    public Node Next { get;set; }

    // Constructor to initialize a new node with data
    public Node(int newData) {
        Data = newData;
        Next = null;
    }
}

class GfG {

    // Function to print the singly linked list
    static void printList(Node head) {
      
        // Base condition is when the head is nullptr
        if (head == null) {
            return;
        }

        // Printing the current node data
        Console.Write(head.Data + " ");

        // Moving to the next node
        printList(head.Next);
    }

    static void Main() {
      
        // Create a linked list: 10 -> 20 -> 30 -> 40
        Node head = new Node(10);
        head.Next = new Node(20);
        head.Next.Next = new Node(30);
        head.Next.Next.Next = new Node(40);

        printList(head);
    }
}

10 20 30 40 

Time Complexity: O(n), where n is number of nodes in the linked list.
Space complexity: O(n) because of recursive stack space.

A

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Article Tags :

• Linked List
• DSA
• Linked Lists

Practice Tags :

• Linked List