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void reversePrint(node *n)
{
if(n)
{ reversePrint(n->next);
print(n->content);
}

}

does the job, but function calls incur a lot of stack space

Other way would be to reverse the linked list,print the list and reverse the list again to restore it

/*reversal*/

void reverseList(node ** head)
{
node* curr = *head, *prev = null;

while(curr)
{
nxt_fwd = curr->next;
curr->next = prev;
prev =curr;
curr=nxt_fwd;
}
*head = prev; /*save the new head */

}

node * prevoius = NULL;
node * current = NULL;
node * following = NULL;

prevoius = head;
current = prevoius->next;
following = current->next;

prevoius->next = NULL;

while(following != NULL)
{

current->next = prevoius;

prevoius = current;
current = following;
following = following->next;

if(following == NULL)
{
current->next = prevoius;
}
}

head = current;
printList(head);

Can be used stack for the same

traverse the linked list, pushing the pointers to each node in a stack as u go along. then pop the elements of the stack and print the data content of the nodes pointed by each of them.

Struct node
{
int data;
Struct node *next;
};

//call the function below from the main loop and pass the pointer to head of list as aparameter to it.

Print_List(Struct node *head)
{
if (head==null)
return;
else
Print_List(head->next);
printf("%d", head->data);
return;
}

void rprint(node* x)
{
if(x->next!=null)
rprint(x->next);
print(x->val);
}

Aadith is perfectly right

public void printBackwards{
for(int i = listToPrint.size() - 1; i >= 0; i--){
System.out.println(listToPrint.get(i));
}
}

public void printBackwards{
for(int i = listToPrint.size() - 1; i >= 0; i--){
System.out.println(listToPrint.get(i));
}
}