Insert a node in Linked List before a given node

Given a node of Linked List N and a value K, the task is to insert the node with value K in the linked list before the given node N.

Structure of the Node:

C++

// Structure of Node
struct Node {
    int data;
    Node* next;

    // Constructor of Node
    Node(int val, Node* link = 0)
        : data(val), next(link)
    {
    }
};

Java

// Structure of Node
public class Node
{
    public int data;
    public Node next;

    // Constructor of Node
    public Node(int val, Node link = null)
    {
        this.data = val;
        this.next = link;
    }
}

// This code is contributed by divyesh072019

Python3

# Structure of Node
class Node:

    # Constructor of Node
    def __init__(self, val, link = None):

        self.data = val
        self.next = link

# This code is contributed by pratham76

C#

// Structure of Node
public class Node
{
    public int data;
    public Node next;

    // Constructor of Node
    public Node(int val, Node link = null)
    {
        this.data = val;
        this.next = link;
    }
};

// This code is contributed by rutvik_56

Javascript

<script>
// Structure of Node
class Node {
    // Constructor of Node
    constructor(val, link = null) {
         this.data = val;
         this.next = link;
    }
}


// This code is contributed by gauravrajput1
</script>

Output:

5 8 6

In the given problem there might be two cases:

• Given node is the head node.
• Given node is any valid node except the head.

When given Node is the Head Node:

The idea is to create a new node with the given value K. Then the next part of the new node will be updated with the pointer head. And finally, the head will be updated with the new node’s address. Below is the image of the same:

When given Node is any valid node except head node:

The simplest approach is to traverse the given linked list to search the previous node of the given node. Then, create the new node with the given value K.Now, update the next part of the new node with the address of the given node and the next part of the previous node with the address of the new node. Below is an illustration of the approach with the help of image:

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int data;
    Node* next;

    // Constructor of Node
    Node(int val, Node* link = 0)
        : data(val), next(link)
    {
    }
};

// Create a head node
Node* head = new Node(5);

// Function prints the linked list
// starting from the given node
void printList(Node* n)
{
    // Till n is not NULL
    while (n != NULL) {

        // Print the data
        cout << n->data << " ";
        n = n->next;
    }
}

// Function to add a node before the
// given node other than head node
void addBefore(Node* given_ptr, int val)
{

    // First check if the given pointer
    // is the address of head
    if (head == given_ptr) {

        // Create a new node
        Node* n = new Node(val);

        // Point to next to current head
        n->next = head;

        // Update the head pointer
        head = n;
        return ;
    }

    // Otherwise traverse the list to
    // find previous node of given node
    else {

        Node *p, *n = head;

        // This loop will return p with
        // previous pointer of given node
        for (n, p; n != given_ptr;
             p = n, n = n->next)
            ;

        // Create a new node
        Node* m = new Node(val);

        // Update the m->next
        m->next = p->next;

        // Update previous node's next
        p->next = m;

    }
}

// Driver Code
int main()
{
    // Head Node
    head->next = new Node(6);

    // Function Call
    addBefore(head->next, 8);

    // Print the linked List
    printList(head);
}

Java

// Java implementation of above approach

import java.io.*;

public class GFG {
    static class Node {
        public int data;
        public Node next;

        // Constructor of Node
        public Node(int val) {
            this.data = val;
            this.next = null;
        }
    }

    static Node head = new Node(5);

    // Function prints the linked list
    // starting from the given node
    static void printList(Node n) {
        // Till n is not null
        while (n != null) {
            // Print the data
            System.out.print(n.data + " ");
            n = n.next;
        }
    }

    // Function to add a node before the
    // given node other than head node
    static void addBefore(Node given_ptr, int val) {
        // First check if the given
        // pointer is the address of
        // head
        if (head == given_ptr) {
            // Create a new node
            Node n = new Node(val);

            // Point to next to current
            // head
            n.next = head;

            // Update the head pointer
            head = n;
            return;
        }

        // Otherwise traverse the list
        // to find previous node of
        // given node
        else {
            Node p = null;

            // This loop will return p with
            // previous pointer of given node
            for (Node n = head; n != given_ptr; p = n, n = n.next);

            // Create a new node
            Node m = new Node(val);

            // Update the m.next
            m.next = p.next;

            // Update previous node's next
            p.next = m;

        }
    }

    // Driver Code
    public static void main(String[] args) {
        // Head Node
        head.next = new Node(6);

        // Function Call
        addBefore(head.next, 8);

        // Print the linked List
        printList(head);
    }
}

Python3

# Python3 program for the above approach
class Node:

    # Constructor of Node
    def __init__(self, val, link = None):
        self.data = val
        self.next = link

# Create a head node
head = Node(5);

# Function prints the linked list
# starting from the given node
def printList(n):

    # Till n is not NULL
    while (n != None):

        # Print the data
        print(n.data, end = ' ')
        n = n.next;

# Function to add a node before the
# given node other than head node
def addBefore(given_ptr, val):
    global head

    # First check if the given pointer
    # is the address of head
    if (head == given_ptr):

        # Create a node
        n = Node(val);

        # Point to next to current head
        n.next = head;

        # Update the head pointer
        head = n;
        return;

    # Otherwise traverse the list to
    # find previous node of given node
    else:

        p = None
        n = head;

        # This loop will return p with
        # previous pointer of given node
        while(n != given_ptr):
            p = n
            n = n.next

        # Create a node
        m = Node(val);

        # Update the m.next
        m.next = p.next;

        # Update previous node's next
        p.next = m;

# Driver Code
if __name__=='__main__':

    # Head Node
    head.next = Node(6);

    # Function Call
    addBefore(head.next, 8);

    # Print the linked List
    printList(head);

    # This code is contributed by rutvik_56

C#

// C# program for the
// above approach
using System;
class GFG{

class Node
{
  public int data;
  public Node next;

  // Constructor of Node
  public Node(int val)
  {
    this.data = val;
    this.next = null;
  }
}

static Node head = new Node(5);

// Function prints the linked list
// starting from the given node
static void printList(Node n)
{
  // Till n is not null
  while (n != null)
  {
    // Print the data
    Console.Write(n.data + " ");
    n = n.next;
  }
}

// Function to add a node before the
// given node other than head node
static void addBefore(Node given_ptr,
                      int val)
{
  // First check if the given
  // pointer is the address of
  // head
  if (head == given_ptr)
  {
    // Create a new node
    Node n = new Node(val);

    // Point to next to current
    // head
    n.next = head;

    // Update the head pointer
    head = n;
    return;
  }

  // Otherwise traverse the list
  // to find previous node of
  // given node
  else
  {
    Node p = null;

    // This loop will return p with
    // previous pointer of given node
    for(Node n = head; n != given_ptr;
        p = n, n = n.next);

    // Create a new node
    Node m = new Node(val);

    // Update the m.next
    m.next = p.next;

    // Update previous node's next
    p.next = m;

  }
}

// Driver Code
public static void Main(String[] args)
{
  // Head Node
  head.next = new Node(6);

  // Function Call
  addBefore(head.next, 8);

  // Print the linked List
  printList(head);
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// Javascript program for the above approach

     class Node {

        // Constructor of Node

    constructor(val) {
        this.data = val;
        this.next = null;
    }

}
    var head = new Node(5);

    // Function prints the linked list
    // starting from the given node
    function printList(n) {

        // Till n is not null
        while (n != null) {

            // Print the data
            document.write(n.data + " ");
            n = n.next;
        }
    }

    // Function to add a node before the
    // given node other than head node
    function addBefore(given_ptr , val) {

        // First check if the given pointer
        // is the address of head
        if (head == given_ptr) {

            // Create a new node
    var n = new Node(val);

            // Point to next to current head
            n.next = head;

            // Update the head pointer
            head = n;
            return;
        }

        // Otherwise traverse the list to
        // find previous node of given node
        else {
    var p = null;

            // This loop will return p with
            // previous pointer of given node
            for (n = head; n != given_ptr; p = n, n = n.next)
                ;

            // Create a new node
    var m = new Node(val);

            // Update the m.next
            m.next = p.next;

            // Update previous node's next
            p.next = m;

        }
    }

    // Driver Code


        // Head Node
        head.next = new Node(6);

        // Function Call
        addBefore(head.next, 8);

        // Print the linked List
        printList(head);

// This code is contributed by todaysgaurav

</script>

Output:

5 8 6

Time Complexity: O(N) 
Auxiliary Space: O(1)