rbdelete.txt
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Properties of a rbtree
======================
1. A node is either red or black.
2. The root is black. (This rule is sometimes omitted. Since the root can
always be changed from red to black, but not necessarily vice-versa,
this rule has little effect on analysis.)
3. All leaves (NIL) are black. (All leaves are same color as the
root.)
4. Every red node must have two black child nodes.
5. Every simple path from a given node to any of its
descendant leaves contains the same number of black nodes.
Assumptions in addition to wikipedia
====================================
lets assume that NULL pointer are black B nodes. But we mark them with a (N)
Example
=======
we start with the tree from the insert example
----------------------------------------------
B(11)
R(8) R(13)
B(3) B(9) B(12) B(16)
B(N) B(N) B(N) R(10) B(N) B(N) B(N) B(N)
B(N) B(N)
we remove R(10) / remove a red node (replace with its child):
wikipedia explains why this can only happen with two leaf children (our B(N))
B(11)
R(8) R(13)
B(3) B(9) B(12) B(16)
B(N) B(N) B(N) B(N) B(N) B(N) B(N) B(N)
again start with the insert example result
------------------------------------------
B(11)
R(8) R(13)
B(3) B(9) B(12) B(16)
B(N) B(N) B(N) R(10) B(N) B(N) B(N) B(N)
B(N) B(N)
remove B(9) (which is the second simple case described on wikipedia)
M black, C red
After remove just repaint child black. As I do not replace the node,
but the value this is simplified in my case to simply do nothing after
normal delete... :D
B(11)
R(8) R(13)
B(3) B(10) B(12) B(16)
B(N) B(N) B(N) B(N) B(N) B(N) B(N) B(N)
again start with the insert example result
------------------------------------------
B(11)
R(8) R(13)
B(3) B(9) B(12) B(16)
B(N) B(N) B(N) R(10) B(N) B(N) B(N) B(N)
B(N) B(N)
now lets delete B(3)... which is stated on wikipedia as the complicated case
where 6 subcases could be distinguished.
Wikipedia says we begin with replacing B(3) which one if its childs, in my
case this means, setting r(8)->left to NULL....
B(11)
R(8) R(13)
B(N) B(9) B(12) B(16)
B(N) R(10) B(N) B(N) B(N) B(N)
B(N) B(N)
So, what is called in on Wikipedia is simply a nullpointer for me...hopefully
I don't have to do anything with it.
Get an overview over our variables now:
N : Nullpointer set in R(8)->left (thus N will be black by definition, ever)
P : R(8)
S : B(9)
Sl: Nullpointer
Sr: R(10)
cases:
- case 2: S is red => reverse color of P and S, then rotate left at P
- case 3: P, S and S's children are black => repaint S red
- case 4: S and S's children are black, P is red => exchange colors of S and
P
- case 5: S and Sr black, Sl is red N left of P => rotate right at S, exchange
colors of S and Sl
- case 6: S black, Sr is red N left of P => rotate left at P, exchange colors
of P and S and make Sr black
looks like case 6:
first rotate left at P
B(11)
B(9) R(13)
R(8) R(10) B(12) B(16)
B(N) B(N) B(N) B(N) B(N) B(N) B(N) B(N)
exchange colors of P and S and make Sr black.
B(11)
R(9) R(13)
B(8) B(10) B(12) B(16)
B(N) B(N) B(N) B(N) B(N) B(N) B(N) B(N)
now for case 2, we delete 16 from the following tree
B(13)
R(8) B(16)
B(3) B(11) B(N) B(N)
B(N) B(N) R(9) B(N)
B(N) B(N)
again first lets see what we have where...
N : Nullpointer set in R(13)->right (thus N will be black by definition, ever)
P : B(13)
S : R(8)
Sl: B(3)
Sr: B(11)
B(13)P
R(8)S B(N)N
B(3)Sl B(11)Sr
B(N) B(N) R(9) B(N)
B(N) B(N)
revert colors of P and S
R(13)P
B(8)S B(N)N
B(3)Sl B(11)Sr
B(N) B(N) R(9) B(N)
B(N) B(N)
rotate right at P
B(8)S
B(3)Sl R(13)P
B(N) B(N) B(11)Sr B(N)N
R(9) B(N)
B(N) B(N)
relable ... done on wikipedia (don't know if I will need it.)
B(8)
B(3) R(13)P
B(N) B(N) B(11)S B(N)N
R(9)Sl B(N)Sr
B(N) B(N)
ok, not case 3... P is red
nor is it case 4 Sl is red.
kind of case 6 reversed ... lets try what they do in the code not in the description...
Sl to black
B(8)
B(3) R(13)P
B(N) B(N) B(11)S B(N)N
B(9)Sl B(N)Sr
B(N) B(N)
rotate right on P
B(8)
B(3) B(11)S
B(N) B(N) B(9)Sl R(13)P
B(N) B(N) B(N)Sr B(N)N
This result is wrong....the balance is ok, but the color of 9 is wrong.
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