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Print the Alternate Nodes in a Linked List without using Recursion



Print the Alternate Nodes in a Linked List without using Recursion

This C program, using iteration, displays the alternate nodes in a linked list. A linked list is an ordered set of data elements, each containing a link to its successor.

    #include <stdio.h>

    #include <stdlib.h>

    struct node

    {

        int a;

        struct node *next;

    };

    void generate(struct node **);

    void display(struct node *);

    void delete(struct node **);

    int main()

    {

        struct node *head = NULL;

        generate(&head);

        printf("\nDisplaying the alternate nodes\n");

        display(head);

        delete(&head);

        return 0;

    }

    void display(struct node *head)

    {

        int flag = 0;

        while(head != NULL)

        {

            if (!(flag % 2))

            {

               printf("%d  ", head->a);

            }

            flag++;

            head = head->next;

        }

    }

    void generate(struct node **head)

    {

        int number, j;

        struct node *temp;

        printf("Enter length of list: ");

        scanf("%d", &number);

        for (j = number; j > 0; j--)

        {

            temp = (struct node *)malloc(sizeof(struct node));

            temp->a = j;

            if (*head == NULL)

            {

                *head = temp;

                (*head)->next = NULL;

            }

            else

            {

                temp->next = *head;

                *head = temp;

            }

        }

    }

    void delete(struct node **head)

    {

        struct node *temp;

        while (*head != NULL)

        {

            temp = *head;

            *head = (*head)->next;

            free(temp);

        }

    }
}