// there's only the root tree
verify(consumer, times(1)).tree(any(Node.class), any(Node.class));
// there's only one feature on the right tree, so all right trees features fall on a single
// bucket
final int leftBucketCount = right.buckets().get().size();
final int expectedBucketCalls = leftBucketCount - 1;
verify(consumer, times(expectedBucketCalls)).bucket(anyInt(), eq(0), any(Bucket.class),
any(Bucket.class));
verify(consumer, times(rightsize - 1)).feature((Node) isNull(), any(Node.class));