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

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: 

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).