In __rb_erase_color(), we have to select one of 3 cases depending on the
color on the 'other' node children.  If both children are black, we flip a
few node colors and iterate.  Otherwise, we do either one or two tree
rotations, depending on the color of the 'other' child opposite to 'node',
and then we are done.

The corresponding logic had duplicate checks for the color of the 'other'
child opposite to 'node'.  It was checking it first to determine if both
children are black, and then to determine how many tree rotations are
required.  Rearrange the logic to avoid that extra check.
Signed-off-by: Michel Lespinasse <walken@google.com>
Cc: Andrea Arcangeli <aarcange@redhat.com>
Acked-by: David Woodhouse <David.Woodhouse@intel.com>
Cc: Rik van Riel <riel@redhat.com>
Cc: Peter Zijlstra <a.p.zijlstra@chello.nl>
Cc: Daniel Santos <daniel.santos@pobox.com>
Cc: Jens Axboe <axboe@kernel.dk>
Cc: "Eric W. Biederman" <ebiederm@xmission.com>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>