**Problem Statement:** Given a node and a doubly linked list, insert the node before the tail of the doubly linked list and return the updated list.

##
**
Examples
**

Example 1:

Input Format:DLL: 1 <-> 2 <-> 3

Value to be Inserted: 4

Result: DLL: 1 <-> 2 <-> 4 <->3

Explanation: We need to insert the value 4 before the tail of the given Doubly Linked List, which is node 3.

Example 2:

Input Format:DLL: 5

Value to be Inserted: 7

Result: DLL: 7 <-> 5

Explanation: In this case, the DLL initially contains only one node with a value of 5. We insert the value 7 before the tail of the DLL, resulting in a DLL with two nodes.

**Solution**

** Disclaimer**:

*Don’t jump directly to the solution, try it out yourself first.*

**Approach**:

To insert a new node before the tail of a Doubly Linked List (DLL), **traverse **the list until **reaching the tail**, while also** tracking the node before it**. Create a new node with the given value, set its **next** to the tail and **back** to the previous node. Update the previous node’s next to the new node and the tail’s back to maintain the doubly linked structure.

An edge case to be mindful of is if there is **only one node** in the entire doubly linked list then the **head would also serve as the tail**. In this case, the steps to insert a node before the head of a doubly linked list can be done.

**Algorithm:**

**Step 1:** Traverse to the tail of the linked list and keep track of it using the **tailpointer**. Also, keep track of the node previous to the tail using the** prev pointer.**

**Step 2:** Create a new node with the given value. Set its **next pointer to point to the tail **and its** back pointer to point to the prev node.**

**Step 3:** Update the **next pointer **of the** prev node** to point to the new node** **and** **the** back pointer **of the** tail** to point to the new node, effectively inserting the new node before the tail.

**Code:**

## C++ Code

```
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
// Define a Node class for doubly linked list
class Node {
public:
// Data stored in the node
int data;
// Pointer to the next node in the list (forward direction)
Node* next;
// Pointer to the previous node in the list (backward direction)
Node* back;
// Constructor for a Node with both data, a reference to the next node, and a reference to the previous node
Node(int data1, Node* next1, Node* back1) {
data = data1;
next = next1;
back = back1;
}
// Constructor for a Node with data, and no references to the next and previous nodes (end of the list)
Node(int data1) {
data = data1;
next = nullptr;
back = nullptr;
}
};
// Function to convert an array to a doubly linked list
Node* convertArr2DLL(vector<int> arr) {
// Create the head node with the first element of the array
Node* head = new Node(arr[0]);
// Initialize 'prev' to the head node
Node* prev = head;
for (int i = 1; i < arr.size(); i++) {
// Create a new node with data from the array and set its 'back' pointer to the previous node
Node* temp = new Node(arr[i], nullptr, prev);
// Update the 'next' pointer of the previous node to point to the new node
prev->next = temp;
// Move 'prev' to the newly created node for the next iteration
prev = temp;
}
// Return the head of the doubly linked list
return head;
}
// Function to print the elements of the doubly linked list
void print(Node* head) {
while (head != nullptr) {
// Print the data in the current node
cout << head->data << " ";
// Move to the next node
head = head->next;
}
}
// This function was explain in prevoius articles
// Please refer there
// Function to insert new node before head
Node* insertBeforeHead(Node* head, int val){
// Create new node with data as val
Node* newHead = new Node(val , head, nullptr);
head->back = newHead;
return newHead;
}
Node* insertBeforeTail(Node* head, int val){
// Edge case, if dll has only one elements
if(head->next==NULL){
// If only one element
return insertBeforeHead(head, val);
}
// Create pointer tail to traverse to the end of list
Node* tail = head;
while(tail->next!=NULL){
tail = tail->next;
}
// Keep track of node before tail using prev
Node* prev = tail->back;
// Create new node with value val
Node* newNode = new Node(val, tail, prev);
// Join the new node into the doubly linked list
prev->next = newNode;
tail->back = newNode;
// Return the updated linked list
return head;
}
int main() {
vector<int> arr = {12, 5, 8, 7, 4};
Node* head = convertArr2DLL(arr);
print(head);
cout << endl << "Doubly Linked List After Inserting 6 before tail: " << endl;
head = insertBeforeTail(head, 6);
print(head);
return 0;
}
```

**Output:**

12 5 8 7 4

Doubly Linked List After Inserting 6 before tail:

12 5 8 7 6 4

**Time Complexity: O(n)** The time complexity of this insertion operation is O(n) because we have to traverse the entire list to reach i’s tail. The complexity would be O(1) if we were given the tail node to start with.

**Space Complexity: O(1) ** The space complexity is also O(1) because a constant amount of extra space is used to create the new node.

## Java Code

```
public class DLinkedList {
public static class Node {
// Data stored in the node
public int data;
// Reference to the next node in the list (forward direction)
public Node next;
// Reference to the previous node in the list (backward direction)
public Node back;
// Constructor for a Node with both data, a reference to the next node, and a reference to the previous node
public Node(int data1, Node next1, Node back1) {
data = data1;
next = next1;
back = back1;
}
// Constructor for a Node with data, and no references to the next and previous nodes (end of the list)
public Node(int data1) {
data = data1;
next = null;
back = null;
}
}
private static Node convertArr2DLL(int[] arr) {
// Create the head node with the first element of the array
Node head = new Node(arr[0]);
// Initialize 'prev' to the head node
Node prev = head;
for (int i = 1; i < arr.length; i++) {
// Create a new node with data from the array and set its 'back' pointer to the previous node
Node temp = new Node(arr[i], null, prev);
// Update the 'next' pointer of the previous node to point to the new node
prev.next = temp;
// Move 'prev' to the newly created node for the next iteration
prev = temp;
}
// Return the head of the doubly linked list
return head;
}
private static void print(Node head) {
while (head != null) {
// Print the data in the current node
System.out.print(head.data + " ");
// Move to the next node
head = head.next; // Move to the next node
}
System.out.println();
}
private static Node insertBeforeHead(Node head, int val){
// Create new node with data as val
Node newHead = new Node(val, head, null);
// Set old head's back pointer to the new Head
head.back = newHead;
// Return the new head
return newHead;
}
private static Node insertBeforeTail(Node head, int val){
// Edge case, if dll has only one elements
if(head.next==null){
// If only one element
return insertBeforeHead(head, val);
}
// Create pointer tail to traverse to the end of list
Node tail = head;
while(tail.next!=null){
tail = tail.next;
}
// Keep track of node before tail using prev
Node prev = tail.back;
// Create new node with value val
Node newNode = new Node(val, tail, prev);
// Join the new node into the doubly linked list
prev.next = newNode;
tail.back = newNode;
// Return the updated linked list
return head;
}
public static void main(String[] args) {
int[] arr = {12, 5, 6, 8, 4};
// Convert the array to a doubly linked list
Node head = convertArr2DLL(arr);
// Print the doubly linked list
print(head);
System.out.println("Doubly Linked List After Inserting 6 before tail: ");
head = insertBeforeTail(head, 6);
print(head);
}
}
```

**Output:**

12 5 8 7 4

Doubly Linked List After Inserting 6 before tail:

12 5 8 7 6 4

**Time Complexity: O(n)** The time complexity of this insertion operation is O(n) because we have to traverse the entire list to reach i’s tail. The complexity would be O(1) if we were given the tail node to start with.

**Space Complexity: O(1) ** The space complexity is also O(1) because a constant amount of extra space is used to create the new node.

## Python Code

```
# Define a Node class for a doubly linked list
class Node:
def __init__(self, data, next_node=None, back_node=None):
# Data stored in the node
self.data = data
# Reference to the next node in the list (forward direction)
self.next = next_node
# Reference to the previous node in the list (backward direction)
self.back = back_node
# Function to convert an array to a doubly linked list
def convert_arr_to_dll(arr):
# Create the head node with the first element of the array
head = Node(arr[0])
# Initialize 'prev' to the head node
prev = head
for i in range(1, len(arr)):
# Create a new node with data from the array and set its 'back' pointer to the previous node
temp = Node(arr[i], None, prev)
# Update the 'next' pointer of the previous node to point to the new node
prev.next = temp
# Move 'prev' to the newly created node for the next iteration
prev = temp
# Return the head of the doubly linked list
return head
# Function to print the elements of the doubly linked list
def print_dll(head):
while head is not None:
# Print the data in the current node
print(head.data, end=" ")
# Move to the next node
head = head.next
print()
# Function to insert a new node before the head of the doubly linked list
def insert_before_head(head, val):
# Create a new node with data as val
new_head = Node(val, head, None)
head.back = new_head
return new_head
# Function to insert a new node before the tail of the doubly linked list
def insert_before_tail(head, val):
# Edge case: if the DLL has only one element
if head.next is None:
# If only one element, insert before the head
return insert_before_head(head, val)
# Create a pointer 'tail' to traverse to the end of the list
tail = head
while tail.next is not None:
tail = tail.next
# Keep track of the node before the tail using 'prev'
prev = tail.back
# Create a new node with value 'val'
new_node = Node(val, tail, prev)
# Join the new node into the doubly linked list
prev.next = new_node
tail.back = new_node
# Return the updated linked list
return head
# Main function
if __name__ == "__main__":
arr = [12, 5, 8, 7, 4]
head = convert_arr_to_dll(arr)
print_dll(head)
print("Doubly Linked List After Inserting 6 before tail:")
head = insert_before_tail(head, 6)
print_dll(head)
```

**Output:**

12 5 8 7 4

Doubly Linked List After Inserting 6 before tail:

12 5 8 7 6 4

**Time Complexity: O(n)** The time complexity of this insertion operation is O(n) because we have to traverse the entire list to reach i’s tail. The complexity would be O(1) if we were given the tail node to start with.

**Space Complexity: O(1) ** The space complexity is also O(1) because a constant amount of extra space is used to create the new node.

## JavaScript Code

```
// Define a Node class for doubly linked list
class Node {
constructor(data, next = null, back = null) {
// Data stored in the node
this.data = data;
// Pointer to the next node in the list (forward direction)
this.next = next;
// Pointer to the previous node in the list (backward direction)
this.back = back;
}
}
// Function to convert an array to a doubly linked list
function convertArr2DLL(arr) {
// Create the head node with the first element of the array
let head = new Node(arr[0]);
// Initialize 'prev' to the head node
let prev = head;
for (let i = 1; i < arr.length; i++) {
// Create a new node with data from the array and set its 'back' pointer to the previous node
let temp = new Node(arr[i], null, prev);
// Update the 'next' pointer of the previous node to point to the new node
prev.next = temp;
// Move 'prev' to the newly created node for the next iteration
prev = temp;
}
// Return the head of the doubly linked list
return head;
}
// Function to print the elements of the doubly linked list
function print(head) {
while (head !== null) {
// Print the data in the current node
console.log(head.data + " ");
// Move to the next node
head = head.next;
}
}
// Function to insert a new node before the head
function insertBeforeHead(head, val) {
// Create a new node with data as val
let newHead = new Node(val, head, null);
head.back = newHead;
return newHead;
}
// Function to insert a new node before the tail
function insertBeforeTail(head, val) {
// Edge case: if DLL has only one element
if (head.next === null) {
// If only one element, insert before the head
return insertBeforeHead(head, val);
}
// Create a pointer 'tail' to traverse to the end of the list
let tail = head;
while (tail.next !== null) {
tail = tail.next;
}
// Keep track of the node before 'tail' using 'prev'
let prev = tail.back;
// Create a new node with value 'val'
let newNode = new Node(val, tail, prev);
// Join the new node into the doubly linked list
prev.next = newNode;
tail.back = newNode;
// Return the updated linked list
return head;
}
const arr = [12, 5, 8, 7, 4];
let head = convertArr2DLL(arr);
print(head);
console.log("\nDoubly Linked List After Inserting 6 before tail: ");
head = insertBeforeTail(head, 6);
print(head);
```

**Output:**

12 5 8 7 4

Doubly Linked List After Inserting 6 before tail:

12 5 8 7 6 4

**Time Complexity: O(n)** The time complexity of this insertion operation is O(n) because we have to traverse the entire list to reach i’s tail. The complexity would be O(1) if we were given the tail node to start with.

**Space Complexity: O(1) ** The space complexity is also O(1) because a constant amount of extra space is used to create the new node.

In case you are learning DSA, you should definitely check out our free A2Z DSA Course with videos and blogs.

Special thanks toGauri Tomarfor contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam,please check out this article