Print in Reverse - Hacker Rank Solution

You are given the pointer to the head node of a linked list and you need to print all its elements in reverse order from tail to head, one element per line. The head pointer may be null meaning that the list is empty - in that case, do not print anything!

Input Format
You have to complete the void ReversePrint(Node* head) method which takes one argument - the head of the linked list. You should NOT read any input from stdin/console.
Output Format
Print the elements of the linked list in reverse order to stdout/console (using printf or cout) , one per line.
Sample Input
1 --> 2 --> NULL
2 --> 1 --> 4 --> 5 --> NULL
Sample Output

2
1
5
4
1
2

Explanation
1. First list is printed from tail to head hence 2,1
2. Similarly second list is also printed from tail to head.
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 /*
  Print elements of a linked list in reverse order as standard output
  head pointer could be NULL as well for empty list
  Node is defined as
  struct Node
  {
     int data;
     struct Node *next;
  }
*/
struct Node *temp1,*temp2,*temp;
void ReversePrint(Node *head)
{
    if (head == NULL) return;

    ReversePrint(head -> next);
    cout << head -> data << endl;
   
   
  // This is a "method-only" submission.
  // You only need to complete this method.
}
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