Get Node Value - Hacker Rank Solution

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.
Constraints
Position will be a valid element in linked list.
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

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 /*
  Get Nth element from the end in a linked list of integers
  Number of elements in the list will always be greater than N.
  Node is defined as 
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
int GetNode(Node *head,int positionFromTail)
{
    int i=0;
    Node* current = head;
    Node* result = head;
    while(current!=NULL)
    {
        current=current->next;
        if (i++>positionFromTail)
        {
            result=result->next;
        }
    }
    return result->data;
    
  // This is a "method-only" submission. 
  // You only need to complete this method. 
}
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