RedBlack Trees

1
Red-Black Trees

• ??????? ???
• 0004425
• ???
• superman_at_it.soongsil.ac.kr

2
Introduction

• In binary search tree,
• If its height is large, their performance may be
  no good
• Red-Black Trees are,
• Balance in order to guarantee that basic
  dynamic-set operations take O(lgn) time in the
  worst case

3
Properties of RB trees

• RB tree has one extra bit of storage per node
• key, left, right, p and
• color, which can be either RED or BLACK
• The path length is more than twice, so tree is
  approximately balanced

4
Properties of RB trees

• External node
• If a child or the parent of a node dose not
  exist, the pointer field of the node contains the
  value NIL
• leave
• Internal node
• key-bearing nodes

5
Properties of RB trees

• A binary search tree is a RB tree if it satisfies
  the following properties
• 1. Every node is either red or black
• 2. Every leaf(NIL) is black
• 3. If a node is red, then both it children are
  black
• 4. Every simple path from a node to descendant
  leaf contains the same number of black nodes

6
Example of a RB tree
3
26
3
17
41
2
30
14
21
47
2
2
1
2
10
16
19
23
28
38
2
1
1
1
1
1
NIL
NIL
1
7
15
35
39
20
12
1
1
1
NIL
NIL
NIL
NIL
NIL
NIL
1
3
1
NIL
NIL
NIL
NIL
NIL
NIL
NIL
NIL
NIL
NIL
NIL
NIL
NIL
7
A RB trees height

• Black-height bh(x)
• Number of black nodes on any path
• In property 4, all descending paths from the node
  have a same number of black nodes

8
A RB trees height

• A RB tree has height at most 2ln(n1)

• Subtree rooted at any node x contains at least
  2bh(x)-1
• If a height of x is 0, then x must be a leaf, and
  2bh(x)-10 internal nodes.
• Consider a node x(positive height, two children).
  
• Each child has a black-height(bh(x) or bh(x)-1)
• Since height of a child of x is less than height
  of x, Hypothesis to conclude that each child has
  at least 2bh(x)-1-1 internal nodes
• The subtree rooted at x consist at least
• (2bh(x)-1-1)(2bh(x)-1-1)1 2bh(x)-1
• Let h be the height of tree
• according to property 3, at least half nodes on
  any simple path(not including root), must be
  black
• Consequently, the black-height of the root must
  be at least h/2
• Thus, n?2h/2-1
• Taking logarithms on both sides lg(n1)?h/2, or
  h?1lg(n1)

9
Complexity of RB tree

• In this lemma
• The dynamic-set operations can be implemented in
  O(lgn) time on RB tree

10
Rotations

• Why we use Rotations
• The TREE_INSERT and TREE_DELETE takes O(lnn), but
  its violate properties
• To restore these properties, we must change the
  colors of the nodes and also change the pointer
  structure
• We change the pointer structure through rotation
• local operation
• preserves the inorder key ordering

11
Rotations

• When Left-Rotation on a node x, its right child y
  is non NIL
• When rotate left, y is the new root, with x as
  ys left child and ys left child as xs right
  child

y
x
Right-Rotate(T,y)
?
?
x
y
?
?
?
?
Left-Rotate(T,x)
12
Left-Rotate(T,x)

• Both rotation operations run in O(1)

• Left-Rotate(T,x) Assumes that rightx?NIL
• 1 y?rightx
• 2 rightx?lefty
• 3 if lefty?NIL
• then plefty?x
• 5 py?px
• 6 if pxNIL
• then rootT?y
• else if xleftpx
• then leftpx?y
• else rightpx?y
• 11 lefty?x
• 12 px?y

13
Left-Rotate(T,x)
7
x
11
4
y
6
18
9
3
19
14
2
22
17
12
Left-Rotate(T,x)
20
7
y
18
4
x
6
19
11
3
22
2
14
9
20
17
12
14
Insertion

• Insertion in RB tree can be in O(lgn)
• when insert node x
• Use the Tree-Insert to insert node x into tree T
  as if it were an ordinary binary search tree
• color x red
• To guarantee the RB properties are preserved,
  recoloring nodes and rotation
• RB-Insert handles the various cases
• Because, it can arise as we fix up the modified
  tree

15
RB-Insert(T,x)

• Three major step
• Determine what violations of properties
• Examine the overall goal of the while loop
• Explore while loop is broken and see how they
  accomplish the goal

16
RB-Insert(T,x)

• RB-Insert(T,x)
• 1 Tree-Insert(T,x)
• 2 colorx?RED
• 3 while x?rootT and colorpxRED
• 4 do if pxleftppx
• then y?rightppx
• if coloryRED
• then colorpx?BLACK
• colory?BLACK
• colorppx?RED
• x?ppx
• else if xrightpx
• then x?px
• Left-Rotate(T,x)
• colorpx?BLACK
• colorppx?RED
• Right-Rotate(T,ppx)
• else(same as then clause with right and
  left exchanged)
• 18 colorrootT?BLACK

17
RB-Insert(T,x)-case 1
new x
C
C
A
A
y
D
D
?
x
?
B
?
B
?
?
?
?
?
?
?
new x
C
C
B
B
y
D
D
?
?
x
?
?
?
A
A
?
?
?
?
?
18
RB-Insert(T,x)-case 2,3
C
y
?
A
x
B
?
?
?
case 2
B
C
y
case 3
?
A
x
C
B
x
?
A
?
?
?
?
?
?
19
RB-Insert(T,x)
11
11
y
14
2
14
2
x
7
15
1
7
15
1
5
8
5
8
y
case1
4
4
x
case2
11
7
y
14
11
x
7
2
x
8
15
2
14
1
5
8
5
4
1
15
case3
4
20
Deletion

• it takes O(lgn) too
• sentinel-If we could ignore the boundary
  conditions at leaf of the tree
• dummy object that allows us to simplify boundary
  conditions
• sentinel NILT is represents NIL but it has all
  the fields of the other elements(color field is
  black and others are arbitrary values)
• Deletion in RB tree, all NIL are replaced by
  sentinel NILT
• RB-Delete-Fixup is change colors and performs
  rotations to restore properties

21
RB-Delete(T,z)
RB-Delete(T,z) 1 if leftznillT or
rightznillT 2 then y?z 3 else
y?Tree-Successor(z) 4 if lefty?nillT 5
then x?lefty 6 else x?righty 7
px?py 8 if pynillT 9 then
rootT?x 10 else if yleftpy 11
then leftpy?x 12 else rightpy?x 13
if y?z 14 then keyz?keyy(if y has other
fields, copy them, too) 15 if coloryBLACK 16
then RB-Delete-Fixup(T,x) 17 return y
22
RB-Delete(T,z)

• Tree-Delete and RB-Delete are like algorithms
• But all NIL in Tree-Delete have been replaced by
  sentinel NILT
• The test for whether x is NIL in Tree-Delete has
  been removed, and the px?py is performed
  unconditionally in RB-Delete if x is sentinel
  NIL, its parent pointer points to the parent of
  the spliced-out node y
• if y is black, RB-Delete-Fixup is excuted

23
RB-Delete-Fixup(T,z)
RB-Delete-Fixup(T,x) 1 while x?rootT and
colorxBLACK 2 do if xleftpx 3
then w?rightpx 4 if
colorwRED 5 then
colorw?BLACK 6
colorpx?RED 7
Left-Rotate(T,px) 8
w?rightpx 9 if colorleftwBLACK
and colorrightwBLACK 10 then
colorw?RED 11 x?px 12
else if colorrightwBLACK 13
then colorleftw?BLACK 14
colorw?RED 15
Right-Rotate(T,w) 16
w?rightpx 17
colorw?colorpx 18
colorpx?BLACK 19
colorrightw?BLACK 20
Left-Rotate(T,px) 21
x?rootT 22 else (same as then clause with
right and left exchanged) 23 colorx?BLACK
24
RB-Delete(T,z)
D
B
E
case 1
new w
A
x
C
?
?
?
?
?
?
new x
c
c
B
B
w
D
D
x
A
A
case 2
E
C
E
C
?
?
?
?
?
?
?
?
?
?
?
?
25
RB-Delete(T,z)
c
c
B
B
x
w
A
new w
D
x
C
A
case 3
?
?
?
E
C
?
?
D
?
?
?
?
?
E
?
?
c
c
B
D
w
D
x
B
A
case 4
E
c
c
E
C
A
C
?
?
?
?
?
?
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new x rootT