**Print in Reverse**

### Problem Statement :

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: SinglyLinkedListNode pointer head: a reference to the head of the list Prints The data values of each node in the reversed list. Input Format: The first line of input contains t, the number of test cases. The input of each test case is as follows: 1. The first line contains an integer n, the number of elements in the list. 2. Each of the next n lines contains a data element for a list node. Constraints: 1. 1<=n<=1000 2. 1<=list[i]<=1000

### Solution :

` ````
Solution in C :
In C:
//the following fuction is all that is needed to complete the
//challenge in hackerrank platform.
void reversePrint(SinglyLinkedListNode* head) {
SinglyLinkedListNode* current=head,*next,*prev=NULL;
while(current!=NULL)
{ next=current->next;
current->next=prev;
prev=current;
current=next;
}
head=prev;
SinglyLinkedListNode* temp=head;
while(temp!=NULL)
{
printf("%d\n",temp->data);
temp=temp->next;
}
}
In C++:
//the following fuction is all that is needed to complete the
//challenge in hackerrank platform.
void ReversePrint(Node *head)
{
// This is a "method-only" submission.
// You only need to complete this method.
if(head!=NULL)
{
ReversePrint(head->next);
cout<<head->data<<"\n";
}
}
In Java:
//the following method is all that is needed to complete the
//challenge in hackerrank platform.
static void reversePrint(SinglyLinkedListNode head) {
if(head.next != null) {
reversePrint(head.next);
}
System.out.println(head.data);
}
In Python 3 :
//the following method is all that is needed to complete the
//challenge in hackerrank platform.
def ReversePrint(head):
if head is None:
return
ReversePrint(head.next)
print(head.data)
```

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