提交 60670b80 编写于 作者: M Michel Lespinasse 提交者: Linus Torvalds

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: NMichel Lespinasse <walken@google.com>
Reviewed-by: NRik 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: NAndrew Morton <akpm@linux-foundation.org>
Signed-off-by: NLinus Torvalds <torvalds@linux-foundation.org>
上级 7abc704a
......@@ -368,17 +368,28 @@ static void __rb_erase_color(struct rb_node *node, struct rb_node *parent,
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 @@ void rb_erase(struct rb_node *node, struct rb_root *root)
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);
}
......
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