Reverse a forward linked list in Theta(N) time. (Note no checks on malloc because this is a toy program)

reverse.c

#include <stdio.h>
#include <stdlib.h>


struct node
{
	struct node * next;
	int seq;
};

int main()
{
	/* Reverse a forward only list, N time */
	//create list:
	struct node one; one.seq = 1;
	struct node two; two.seq = 2;
	struct node three; three.seq = 3;
	struct node * head = malloc(sizeof(struct node)); head->next = &one; 
	struct node * clnup = head; //for your memory leak.
	one.next = &two; two.next = &three; three.next = NULL;
	for (head = &one; head != NULL; head = head->next)
		printf("%d\n", head->seq);
	//Traverse
	head = &one;
	struct node * bckPtr = NULL;
	struct node * tmp = NULL;
	struct node * orgHead = head;
	bckPtr = head;
	head = head->next; //prime the pump
	for (; head != NULL; ){
		tmp = head->next;
		head->next = bckPtr;
		bckPtr = head;
		head = tmp;
	}
	//flip the original head.
	orgHead->next = NULL;
	
	for (head = bckPtr; head != NULL; head = head->next)
		printf("%d\n", head->seq);

	free(clnup);

	return 0;
}