Skip to content
Related Articles
Print the last k nodes of the linked list in reverse order | Recursive approach
• Difficulty Level : Medium
• Last Updated : 07 Jun, 2021

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```

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 ``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

 ``
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 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).

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live

My Personal Notes arrow_drop_up