Get Node Value

Problem Statement

You’re given the pointer to the head node of a linked list and a specific position. Counting backwards from the tail node of the linked list, get the value of the node at the given position. A position of 0 corresponds to the tail, 1 corresponds to the node before the tail and so on.

Input Format You have to complete the int GetNode(Node* head, int positionFromTail) method which takes two arguments - the head of the linked list and the position of the node from the tail. positionFromTail will be at least 0 and less than the number of nodes in the list. You should NOT read any input from stdin/console.

Output Format Find the node at the given position counting backwards from the tail. Then return the data contained in this node. Do NOT print anything to stdout/console.

Sample Input 1 -> 3 -> 5 -> 6 -> NULL, positionFromTail = 0 1 -> 3 -> 5 -> 6 -> NULL, positionFromTail = 2

Sample Output 6 3