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We may do it recursively like postorder traversal

Since it is a BST, We need to do a inorder traversal.
And for each node traversed add a link to the previous element as well as the next element.

I agree with him. We should a do inorder traversal to make sorted doubly link list.

how the relationship holds?

Use ordered BST and have two pointers for small and large in BST. While converting use the small pointer as previous and large pointer as next.(small pointer has the address of the node which is lesser than the parent and the large pointer has the address of the node which is greater than the parent.

{
Node * conv(Node *N)
{
static Node *H=NULL;
static Node *p=NULL;

Node *t1=N->lc;
Node *t2=N->rc;

if(t1)
conv(t1);

if(!H)
{
H=N;
p=N;
}
else
{
p->rc=N;
N->lc=p;
N->rc=NULL;
p=N;
}

if(t2)
conv(t2);

return H;
}
}

It should be BFS traversal of the tree and we must add nodes in the list the order in which they are added in the queue during BFS.
Making sorted linked list makes no sense.