CareerCup

easy... just for each parent do a right rotation and at the same time set prev of parent to left child(if any).

do a in-order traversal, with head and tail pointers.

struct node {
	int data;
	node * left;
	node * right;
	node(int a) {
		data=a; left=NULL; right=NULL;
	}
};

node * head=NULL, * tail=NULL;
int cnt;

void bsttodll(node * root) {
	if(root==NULL) return;
	bsttodll(root->left);
	if(head==NULL) {
		head = root;
		tail = head;
	} else {
		root->left = tail;
		tail->right = root;
		tail = root;
	}
	bsttodll(root->right);
	return;
}

int main (int argc, char const* argv[]) {
	
	node * root = new node(7);
	root->left = new node(4);
	root->right = new node(10);
	root->left->left = new node(3);
	root->left->right = new node(5);
	root->right->left = new node(9);
	
	bsttodll(root);
	
	return 0;
}

int the in order traversal of bst...make following changes..
inorder(root,first,last){
if(root==null)
return null;
inorder(root->left,first,last);
if(last==null)
{
last=first=root;
}
else
{
last->right=root;
root->left=last;
last=root;
}
inorder(root->right,first,last);
}

i think it is pretty easy :) just do the inorder traversal and accordingly change
see this --
void BSTtoDLL(struct node *r,struct node **head,struct node *prev)
{
if(!r)
return;
BSTtoDLL(r->left,head,prev);
if(!*head)
{
*head=r;
}
else
{
r->left=prev;
prev->right=r;}

prev=r;

BSTtoDLL(r->right,head,prev);
}

where the head is head of linked list