Bobby, that is not constant space because it uses O(N) stack space. There are obvoius O(N^2)-time O(1)-space algorithms, and obvious O(N) time O(N) space algorithms. This is my best guess. Assuming you have exclusive access to the list, you can reverse it, walk it, and then reverse it again. Something like this: #include #include struct node { int value; struct node * next; }; void print_backwards( node * head ) { node * prev = NULL; node * cur = head; node * next; while( cur ) { next = cur->next; cur->next = prev; prev = cur; cur = next; } cur = prev; prev = NULL; while( cur ) { printf( "%d\n", cur->value ); next = cur->next; cur->next = prev; prev = cur; cur = next; } assert( prev == head ); } main() { node a, b, c; a.value = 1; a.next = &b; b.value = 2; b.next = &c; c.value = 3; c.next = NULL; print_backwards( &a ); }

void BWDisplayLinkedList(node* pHead) { if(!pHead) return; BWDisplayLinkedList(pHead->next); cout data "; }

Does this work well? Reverse a list, print it, and then reverse it again. struct node *reverse(struct node *oldlist) { struct node *newlist = NULL; while(oldlist!=NULL) { struct node *temp = oldlist; oldlist=oldlist->next; temp->next=newlist; newlist=temp; } return newlist; } void display(struct node **q) { struct node *temp; temp = *q; if(*q==NULL) { printf("List is empty\n\n"); } else { while(temp!=NULL) { printf("%d=>",temp->data); temp=temp->next; } printf("||\n\n"); } } //p is our list p = reverse(p); display(&p); p = reverse(p);

Thanks to recursion :) void print_backward(node* n) { if(n == NULL) return; print_backward(n->nxt); cout val << endl; }

Yes recursion does the job in linear and constant time. :-)

void f1(LinkedListNode lln) { if(lln.next != null) f1(lln.next); System.out.print(lln.value); }

One can create a function that takes a node as an argument and checks whether the next of the passed node is NULL or not.In case it is not NULL,the same function is again called for the NEXT node.if the Next of the passed node is NULL,the function prints the value of the node and returns. void f1(node* p) { if(p->next!= NULL) { f1(p->next) } else { print ("%d",p->value); } }