#include #include

typedef struct nod{
int info;
struct nod *next;
}node;

node* SortInsert(node *root, int item); //this function is complete
void simplePrint(node* root); //this function is complete
int countNodes (node* root); //to create this function
node* EvenCopy(node * root); //to create this function

int main()
{
node* head=NULL;
node* head2 = NULL;

node *t;
int ch, ele;
head = SortInsert(head, 4);
head = SortInsert(head,6);
head = SortInsert(head,3);
head = SortInsert(head,5);

printf("\nSimple print List 1: ");
simplePrint(head);

printf("\nCount Nodes %d", countNodes(head)); //modify the countNodes function to make it work

head2 = EvenCopy(head); //modify the EvenCopy function to make it work
printf("\nSimple print after EvenCopy: ");
simplePrint(head2); //This call should print 4, 6

return 0;

}

void simplePrint(node* root)
{
node* t=root;
while(t!=NULL)
{
printf("%d ",t->info);
t=t->next;
}
}

node* SortInsert(node* root, int item)
{
node *temp;
node *t;
temp= (node *) malloc(sizeof(node));
temp->info=item;
temp->next=NULL;
if (root==NULL || root->info >=item)
{
temp->next = root;
root = temp;
}
else
{
t = root;
while (t->next != NULL && t->next->info next;
temp->next = t->next;
t->next = temp;
}

return root;
}

THE FOLLOWING QUESTIONS ARE HERE:

int countNodes (node* root)
{
/*this function takes the head of a linked list and returns the number of nodes available in the linked list. You can use for loops or recursion */

}

node* EvenCopy(node * root)
{
/*this function takes the head of a linked list and copies all the even numbers to a new linked list and return the
head of the new linked list. Note that a number is considered as even if number%2 is 0.
Example: passing the head of a linked list containing 3, 4, 5, 6 would return another linked list containing 4, 6 */

}