Do not short-cut the mutex protecting node->subreq.done

A follow-up issue has been reported in #13:

Assert error in Lck_Delete(), cache/cache_lck.c line 309:
  Condition((pthread_mutex_destroy(&ilck->mtx)) == 0) not true.

triggered from Lck_Delete(&bytes_tree->nodes_lock) at the bottom of
bytes_tree_fini().

Assuming everything else working correctly, the only scenario I
can see the moment is that we see the node->subreq.done == 1
earlier than Lck_Unlock() returned in vped_task(). In this
case, we could advance to destroying the lock while the other
thread still holds it.

The other use case of the shared lock is in fini_final(), where
we already go through an explicit lock/unlock.

Hopefully fixes #13 for real

src/node.c

@@ -622,8 +622,7 @@ fini_subreq(const struct vdp_ctx *vdx, struct node *node)
 
 	assert(node->type == T_SUBREQ);
 
-	if (node->subreq.done == 0)
-		subreq_wait_done(node);
+	subreq_wait_done(node);
 	AN(node->subreq.done);
 
 	if (node->subreq.req == NULL) {
@@ -727,8 +726,7 @@ push_subreq(struct req *req, struct bytes_tree *tree,
 	assert(node->type == T_SUBREQ);
 	(void) next;
 
-	if (node->subreq.done == 0)
-		subreq_wait_done(node);
+	subreq_wait_done(node);
 	AN(node->subreq.done);
 
 	subreq = subreq_fixup(node, req);