Insert Node at the End of a Linked List (original) (raw)

Last Updated : 21 Apr, 2026

Given a head of the **linked list, we need to insert a new node at the end of the linked list.

**Examples:

**Input: x = 6

Output: 1 -> 2 -> 3 -> 4 -> 5 -> 6
Explanation: We can see that 6 is inserted at the end of the linkedlist.**

**Input: x = 1

**Output: 4 -> 5 -> 1
**Explanation: We can see that 1 is inserted at the end of the linked list.

Try It Yourself

**Approach:

Inserting at the end involves traversing the entire list until we reach the last node. We then set the **last node's next reference to point to the new node, making the **new node the last element in the list.

Following is the approach to add a new node at the end of the linked list:

• Create a new node and set its next pointer as NULL since it will be the last node.
• Store the head reference in a temporary variable
• If the Linked List is empty, make the new node as the head and return
• Else traverse till the last node
• Change the next pointer of the last node to point to the new node

Below is the implementation of the approach:

C++ `

#include using namespace std;

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

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

};

// Given the head of a list and an int, appends // a new node at the end and returns the head. Node* insertAtEnd(Node* head, int x) {

// Create a new node
Node* newNode = new Node(x);

// If the Linked List is empty, make
// the new node as the head and return
if (head == nullptr) {
    return newNode;
}

// Store the head reference in a temporary variable
Node* last = head;

// Traverse till the last node
while (last->next != nullptr) {
    last = last->next;
}

// Change the next pointer of the last node 
// to point to the new node
last->next = newNode;

// Return the head of the list
return head;

}

// This function prints the contents // of the linked list starting from the head void printList(Node* node) { while (node != nullptr) { cout << node->data; if (node->next != nullptr) { cout << " -> "; } node = node->next; } cout << endl; }

int main() {

// Create a linked list:
// 1 -> 2 -> 3 -> 4 -> 5
Node* head = new Node(1);
head->next = new Node(2);
head->next->next = new Node(3);
head->next->next->next = new Node(4);
head->next->next->next->next = new Node(5);

head = insertAtEnd(head, 6);

printList(head);

return 0;

}

C

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

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

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

// Function to insert a new node at the end of the list struct Node *insertAtEnd(struct Node *head, int x) { struct Node *newNode = createNode(x);

// If the list is empty, return the new node as head
if (head == NULL)
    return newNode;

// Traverse to the last node
struct Node *last = head;
while (last->next != NULL)
{
    last = last->next;
}

// Link the last node to the new node
last->next = newNode;

return head;

}

// Function to print the linked list void printList(struct Node *node) { while (node != NULL) { printf("%d", node->data); if (node->next != NULL) printf(" -> "); node = node->next; } printf("\n"); }

int main() { // Create a linked list: 1 -> 2 -> 3 -> 4 -> 5 struct Node *head = createNode(1); head->next = createNode(2); head->next->next = createNode(3); head->next->next->next = createNode(4); head->next->next->next->next = createNode(5);

// Insert a new node at the end
head = insertAtEnd(head, 6);

// Print the linked list
printList(head);

return 0;

}

Java

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

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

}

class GfG {

// Given the head of a list and an int, appends
// a new node at the end and returns the head.
static Node insertAtEnd(Node head, int x) {
  
    // Create a new node
    Node newNode = new Node(x);

    // If the Linked List is empty, make
    // the new node as the head and return
    if (head == null) {
        return newNode;
    }

    // Store the head reference in a temporary variable
    Node last = head;

    // Traverse till the last node
    while (last.next != null) {
        last = last.next;
    }

    // Change the next pointer of the last node 
    // to point to the new node
    last.next = newNode;

    // Return the head of the list
    return head;
}

// This function prints the contents 
// of the linked list starting from the head
static void printList(Node node) {
    while (node != null) {
        System.out.print(node.data);
        if (node.next != null) {
            System.out.print(" -> ");
        }
        node = node.next;
    }
    System.out.println();
}

public static void main(String[] args) {
  
    // Create a linked list:
    // 1 -> 2 -> 3 -> 4 -> 5 
    Node head = new Node(1);
    head.next = new Node(2);
    head.next.next = new Node(3);
    head.next.next.next = new Node(4);
    head.next.next.next.next = new Node(5);

    head = insertAtEnd(head, 6);

    printList(head);
}

}

Python

A linked list node

class Node: def init(self, x): # Constructor to initialize a new node with data self.data = x self.next = None

Given the head of a list and an int, appends

a new node at the end and returns the head.

def insertAtEnd(head, x): # Create a new node newNode = Node(x)

# If the Linked List is empty, make
# the new node as the head and return
if head is None:
    return newNode

# Store the head reference in a temporary variable
last = head

# Traverse till the last node
while last.next is not None:
    last = last.next

# Change the next pointer of the last node 
# to point to the new node
last.next = newNode

# Return the head of the list
return head

This function prints the contents

of the linked list starting from the head

def printList(node): while node is not None: print(node.data, end="") if node.next is not None: print(" -> ", end="") node = node.next print()

if name == "main": # Create a linked list: # 1 -> 2 -> 3 -> 4 -> 5 head = Node(1) head.next = Node(2) head.next.next = Node(3) head.next.next.next = Node(4) head.next.next.next.next = Node(5)

head = insertAtEnd(head, 6)

printList(head)

C#

using System;

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

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

}

class GfG {

// Given the head of a list and an int, appends
// a new node at the end and returns the head.
static Node insertAtEnd(Node head, int x) {
    // Create a new node
    Node newNode = new Node(x);

    // If the Linked List is empty, make
    // the new node as the head and return
    if (head == null) {
        return newNode;
    }

    // Store the head reference in a temporary variable
    Node last = head;

    // Traverse till the last node
    while (last.next != null) {
        last = last.next;
    }

    // Change the next pointer of the last node 
    // to point to the new node
    last.next = newNode;

    // Return the head of the list
    return head;
}

// This function prints the contents 
// of the linked list starting from the head
static void printList(Node node) {
    while (node != null) {
        Console.Write(node.data);
        if (node.next != null) {
            Console.Write(" -> ");
        }
        node = node.next;
    }
    Console.WriteLine();
}

static void Main(string[] args) {
    // Create a hard-coded linked list:
    // 1 -> 2 -> 3 -> 4 -> 5 
    Node head = new Node(1);
    head.next = new Node(2);
    head.next.next = new Node(3);
    head.next.next.next = new Node(4);
    head.next.next.next.next = new Node(5);

    head = insertAtEnd(head, 6);

    printList(head);
}

}

JavaScript

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

// Given the head of a list and an int, appends // a new node at the end and returns the head. function insertAtEnd(head, x) { // Create a new node let newNode = new Node(x);

// If the Linked List is empty, make
// the new node as the head and return
if (head === null) {
    return newNode;
}

// Store the head reference in a temporary variable
let last = head;

// Traverse till the last node
while (last.next !== null) {
    last = last.next;
}

// Change the next pointer of the last node 
// to point to the new node
last.next = newNode;

// Return the head of the list
return head;

}

// This function prints the contents // of the linked list starting from the head function printList(node) { while (node !== null) { process.stdout.write(node.data.toString()); if (node.next !== null) { process.stdout.write(" -> "); } node = node.next; } console.log(); }

// Driver code function main() { // Create a hard-coded linked list: // 1 -> 2 -> 3 -> 4 -> 5 let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5);

head = insertAtEnd(head, 6);

printList(head);

}

main();

`

Output

1 -> 2 -> 3 -> 4 -> 5 -> 6

**Time Complexity: O(n) where nis the length of the linked list
**Auxiliary Space: O(1)