rbtree: place easiest case first in rb_erase()
In rb_erase, move the easy case (node to erase has no more than
1 child) first. I feel the code reads easier that way.
Signed-off-by: Michel Lespinasse <walken@google.com>
Reviewed-by: Rik van Riel <riel@redhat.com>
Cc: Peter Zijlstra <a.p.zijlstra@chello.nl>
Cc: Andrea Arcangeli <aarcange@redhat.com>
Cc: David Woodhouse <dwmw2@infradead.org>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
diff --git a/lib/rbtree.c b/lib/rbtree.c
index de89a61..bde1b5c 100644
--- a/lib/rbtree.c
+++ b/lib/rbtree.c
@@ -368,17 +368,28 @@
void rb_erase(struct rb_node *node, struct rb_root *root)
{
- struct rb_node *child, *parent;
+ struct rb_node *child = node->rb_right, *tmp = node->rb_left;
+ struct rb_node *parent;
int color;
- if (!node->rb_left)
- child = node->rb_right;
- else if (!node->rb_right)
- child = node->rb_left;
- else {
+ if (!tmp) {
+ case1:
+ /* Case 1: node to erase has no more than 1 child (easy!) */
+
+ parent = rb_parent(node);
+ color = rb_color(node);
+
+ if (child)
+ rb_set_parent(child, parent);
+ __rb_change_child(node, child, parent, root);
+ } else if (!child) {
+ /* Still case 1, but this time the child is node->rb_left */
+ child = tmp;
+ goto case1;
+ } else {
struct rb_node *old = node, *left;
- node = node->rb_right;
+ node = child;
while ((left = node->rb_left) != NULL)
node = left;
@@ -402,18 +413,8 @@
node->__rb_parent_color = old->__rb_parent_color;
node->rb_left = old->rb_left;
rb_set_parent(old->rb_left, node);
-
- goto color;
}
- parent = rb_parent(node);
- color = rb_color(node);
-
- if (child)
- rb_set_parent(child, parent);
- __rb_change_child(node, child, parent, root);
-
-color:
if (color == RB_BLACK)
__rb_erase_color(child, parent, root);
}