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rb_tree: make clear distinction between two different cases in rb_erase()
There are two cases when a node, having 2 childs, is erased:
'normal case': the successor is not the right-hand-child of the node to be erased
'special case': the successor is the right-hand child of the node to be erased
Here some ascii-art, with following symbols (referring to the code):
O: node to be deleted
N: the successor of O
P: parent of N
C: child of N
L: some other node
normal case:
O N
/ \ / \
/ \ / \
L \ L \
/ \ P ----> / \ P
/ \ / \
/ /
N C
\ / \
\
C
/ \
special case:
O|P N
/ \ / \
/ \ / \
L \ L \
/ \ N ----> / C
\ / \
\
C
/ \
Notice that for the special case we don't have to reconnect C to N.
Signed-off-by: Wolfram Strepp <wstrepp@gmx.de>
Signed-off-by: Peter Zijlstra <a.p.zijlstra@chello.nl>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
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parent
16c047add3
commit
4c60117811
1 changed files with 4 additions and 4 deletions
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@ -244,13 +244,13 @@ void rb_erase(struct rb_node *node, struct rb_root *root)
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parent = rb_parent(node);
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color = rb_color(node);
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if (child)
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rb_set_parent(child, parent);
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if (parent == old) {
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parent->rb_right = child;
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parent = node;
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} else
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} else {
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if (child)
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rb_set_parent(child, parent);
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parent->rb_left = child;
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}
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node->rb_parent_color = old->rb_parent_color;
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node->rb_right = old->rb_right;
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