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Print the last k nodes of the linked list in reverse order | Recursive approach

Last Updated : 11 Jul, 2025
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Given a linked list containing N nodes and a positive integer k should be less than or equal to N. The task is to print the last k nodes of the list in reverse order.

Examples: 

Input: list: 1->2->3->4->5, k = 2                       
Output: 5 4

Input: list: 3->10->6->9->12->2->8, k = 4 
Output: 8 2 12 9

Source: Amazon Interview Experience SDE Off Campus.

Recursive Approach: Recursively traverse the linked list. When returning from each recursive call keep track of the node number, considering the last node as number 1, second last as number 2 and so on. This counting could be tracked with the help of a global or pointer variable. With the help of this count variable, print the nodes having a node number less than or equal to k.

Below is the implementation of the above approach: 

C++ 
// C++ implementation to print the last k nodes
// of linked list in reverse order
#include <bits/stdc++.h>
using namespace std;

// Structure of a node
struct Node {
    int data;
    Node* next;
};

// Function to get a new node
Node* getNode(int data)
{
    // allocate space
    Node* newNode = new Node;

    // put in data
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}

// Function to print the last k nodes
// of linked list in reverse order
void printLastKRev(Node* head,
                     int& count, int k)
{
    // if list is empty
    if (!head)
        return;

    // Recursive call with the next node
    // of the list
    printLastKRev(head->next, count, k);

    // Count variable to keep track of
    // the last k nodes
    count++;

    // Print data
    if (count <= k)
        cout << head->data << " ";
}

// Driver code
int main()
{
    // Create list: 1->2->3->4->5
    Node* head = getNode(1);
    head->next = getNode(2);
    head->next->next = getNode(3);
    head->next->next->next = getNode(4);
    head->next->next->next->next = getNode(5);

    int k = 4, count = 0;

    // print the last k nodes
    printLastKRev(head, count, k);

    return 0;
}
Java 
// Java implementation to print the last k nodes 
// of linked list in reverse order 
class GfG 
{

// Structure of a node 
static class Node 
{ 
    int data; 
    Node next; 
}

// Function to get a new node 
static Node getNode(int data) 
{ 
    // allocate space 
    Node newNode = new Node(); 

    // put in data 
    newNode.data = data; 
    newNode.next = null; 
    return newNode; 
} 

static class C
{
    int count = 0;
}

// Function to print the last k nodes 
// of linked list in reverse order 
static void printLastKRev(Node head, C c, int k) 
{ 
    // if list is empty 
    if (head == null) 
        return; 

    // Recursive call with the next node 
    // of the list 
    printLastKRev(head.next, c, k); 

    // Count variable to keep track of 
    // the last k nodes 
    c.count++; 

    // Print data 
    if (c.count <= k) 
        System.out.print(head.data + " "); 
} 

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

    int k = 4;
    C c = new C();

    // print the last k nodes 
    printLastKRev(head, c, k); 
}
} 
 
// This code is contributed by prerna saini
Python 
# Python implementation to print the last k nodes 
# of linked list in reverse order 

# Node class 
class Node: 
    
    # Function to initialise the node object 
    def __init__(self, data): 
        self.data = data # Assign data 
        self.next =None

# Function to get a new node 
def getNode(data): 

    # allocate space 
    newNode = Node(0) 

    # put in data 
    newNode.data = data 
    newNode.next = None
    return newNode 

class C: 
    def __init__(self, data): 
        self.count = data 

# Function to print the last k nodes 
# of linked list in reverse order 
def printLastKRev(head, c, k): 

    # if list is empty 
    if (head == None): 
        return

    # Recursive call with the next node 
    # of the list 
    printLastKRev(head.next, c, k) 

    # Count variable to keep track of 
    # the last k nodes 
    c.count = c.count + 1

    # Print data 
    if (c.count <= k) :
        print(head.data, end = " ") 

# Driver code 

# Create list: 1->2->3->4->5 
head = getNode(1) 
head.next = getNode(2) 
head.next.next = getNode(3) 
head.next.next.next = getNode(4) 
head.next.next.next.next = getNode(5) 

k = 4
c = C(0)

# print the last k nodes 
printLastKRev(head, c, k) 

# This code is contributed by Arnab Kundu
C# 
// C# implementation to print the last k  
// nodes of linked list in reverse order
using System;

class GFG 
{

// Structure of a node 
public class Node 
{ 
    public int data; 
    public Node next; 
}

// Function to get a new node 
static Node getNode(int data) 
{ 
    // allocate space 
    Node newNode = new Node(); 

    // put in data 
    newNode.data = data; 
    newNode.next = null; 
    return newNode; 
} 

public class C
{
    public int count = 0;
}

// Function to print the last k nodes 
// of linked list in reverse order 
static void printLastKRev(Node head, C c, int k) 
{ 
    // if list is empty 
    if (head == null) 
        return; 

    // Recursive call with the next 
    // node of the list 
    printLastKRev(head.next, c, k); 

    // Count variable to keep track  
    // of the last k nodes 
    c.count++; 

    // Print data 
    if (c.count <= k) 
        Console.Write(head.data + " "); 
} 

// Driver code 
public static void Main(String []args) 
{ 
    
    // Create list: 1->2->3->4->5 
    Node head = getNode(1); 
    head.next = getNode(2); 
    head.next.next = getNode(3); 
    head.next.next.next = getNode(4); 
    head.next.next.next.next = getNode(5); 

    int k = 4;
    C c = new C();

    // print the last k nodes 
    printLastKRev(head, c, k); 
}
} 

// This code is contributed by Arnab Kundu
JavaScript 
<script>

// Javascript implementation of the approach

// Structure of a node
class Node {
        constructor() {
                this.data = 0;
                this.next = null;
             }
        }
        
        
// Function to get a new node
function getNode( data)
{
    // allocate space
    var newNode = new Node();

    // put in data
    newNode.data = data;
    newNode.next = null;
    return newNode;
}

 class C
{
constructor() {
    this.count = 0;
    }
}

// Function to print the last k nodes
// of linked list in reverse order
function printLastKRev( head,  c,  k)
{
    // if list is empty
    if (head == null)
        return;

    // Recursive call with the next node
    // of the list
    printLastKRev(head.next, c, k);

    // Count variable to keep track of
    // the last k nodes
    c.count++;

    // Print data
    if (c.count <= k)
        document.write(head.data + " ");
}

// Driver Code

// Create list: 1->2->3->4->5
var head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);

let k = 4;
let c = new C();

// print the last k nodes
printLastKRev(head, c, k);

// This code is contributed by jana_sayantan.
</script>

Output
5 4 3 2

Time Complexity: O(n).

Iterative Approach: The idea is to use Stack Data Structure. 

  1. Push all linked list nodes to a stack.
  2. Pop k nodes from the stack and print them.

Time Complexity: O(n).

Two Pointer Approach The idea is similar to find k-th node from end of linked list. 

  1. Move first pointer k nodes ahead.
  2. Now start another pointer, second from head.
  3. When first pointer reaches end, second pointer points to k-th node.
  4. Finally using the second pointer, print last k nodes.

Time Complexity: O(n).


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