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>

lib/rbtree.c

@@ -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);
 }