Print reversely the elements of a list using recursion - c

Question

I want to print reversely the elements of a list using recursion, but I get an execution error !
Can someone help?
typedef struct node
{
int val;
struct node* next;
}node;
typedef struct list
{
node* head;
}list;
void display(list L)
{
if(L.head == NULL) return;
L.head= L.head->next;
display(L);
printf("%d\t",L.head->val);
}

Answer 1

void display(node *N){
if(N != NULL){
display(N->next);
printf("%d\t",N->val);
}
}
and in main:
int main (void)
{
//...
display (L.head);

Answer 2

Instead of using :
typedef struct list
{
node* head;
}list;
you can simply pass the "node n" instead, and call it recursively using n->next, as follows:
void display(node *n){
if(n != NULL){
display(n->next);
printf("%d\t",n->val);
}
}
And then call that method from the outside as follows:
display(node->head);
or
display(node.head);
Depending on how you have allocated the structure "node".
Side Note:
I would actually renamed your structures to:
typedef struct List
{
int val;
struct List* next;
}List;
and then call:
void display(List *l){
if(l != NULL){
display(l->next);
printf("%d\t",l->val);
}
}
IMO it better expresses the use-case in hand.
A Full example:
#include <stdio.h>
#include <stdio.h>
#include <stdlib.h>
typedef struct List
{
int val;
struct List* next;
}List;
void display(List *l){
if(l != NULL){
display(l->next);
printf("%d\t",l->val);
}
}
List *addElement(List *l, int val){
if(l == NULL){
List *node = malloc(sizeof(List));
node->val = val;
node->next = NULL;
return node;
}
else
l->next = addElement(l->next, val);
return l;
}
void freeList(List *l){
if(l != NULL){
freeList(l->next);
free(l);
}
}
int main(){
List *l = NULL;
l = addElement(l, 1);
l = addElement(l, 2);
l = addElement(l, 3);
l = addElement(l, 4);
l = addElement(l, 5);
display(l);
freeList(l);
}
OUTPUT:
5 4 3 2 1