red -black tree
DESCRIPTION
red -black tree. Lai Ah Fur. Background: AVL trees may require many restructure operations (rotations) to be performed after an element removal, (2,4) trees may require many fusing or split operations to be performed after either an insertion or removal. - PowerPoint PPT PresentationTRANSCRIPT
red-black tree
Lai Ah Fur
•Background:
•AVL trees may require many restructure operations (rotations) to be performed after an element removal,
•(2,4) trees may require many fusing or split operations to be performed after either an insertion or removal.
•The red-black tree, does not have these drawbacks. It requires that only O(1) structural changes be made after an update in order to stay balanced.
Property of A red-black tree• A red-black tree is a binary search tree with
nodes colored red and black in a way that satisfies the following properties:– Root Property: The root is black.– External Property: Every external node is black.– Internal Property: The children of a red node are
black.– Depth property: All the external nodes have the
same black depth, which is defined as the number of black ancestors minus one.
– The height of a red-black tree storing n items is O(log n).
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4
6 8
7 11
17
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Example of A red-black tree
x
y
β
α
γ
x
y
βα
y
x
βα
γ
LEFT-ROTATE(T,x)
RIGHT-ROTATE(T,y)
LEFT-ROTATE(T,x)1 y←right[x] //Set y.
2 right[x] ← left[y] //Turn y’s left subtree into x’s right subtree.
3 if left[x] ≠ nil[T]
4 then p[left[y] ] ←x // β’s father
5 p[y ] ← p[x ] //Link x’s parent to y.
6 if p[x ] = nil[T]
7 then root [T] ←y
8 else if x=left[p[x]]
9 then left[p[x]] ←y
10 else right[p[x]] ←y
11 left[y] ←x //Put x on y’s left
12 p[x ] ←y
RB-INSERT(T,z) //insert z into T
1 y ← nil[T]
2 x ← root[T]
3 while x≠ nil[T]
4 do y←x
5 if key[z]< key[x]
6 then x ← left[x]
7 else x ← right [x]
8 p[z] ←y
9 if y=nil[T]
10 then root[T] ← z
11 else if key[z]<key[y]
12 then left[y] ← z
13 else right[y] ← z
14 left[z] ←nil[T]
15 right[z] ← nil[T]
16 color[z] ← RED
17 RB-INSERT-FIXUT(T,z)
RB-INSERT-FIXUP(T,z)
1 while color [p[z]] =RED
2 do if p[z]=left [p[p[z]]]
3 then y ← right[p[p[z]]]
4 if color [y] =RED
5 then color p[z] ←BLACK //case 1
6 color [y ] ←BLACK //case 1
7 color [p[p[z]]] ← RED //case 1
8 z ←[p[z]] //case 1
9 else if z = right [p[z]]
10 then z← p[z] //case 2
11 LEFT-ROTATE(T,z) //case 2
12 color [p[z]] ←BLACK //case 3
13 color [p[p[z]]] ← RED // //case 3
14 RIGHT-ROTATE(T, p[p[z]]) //case 3
15 else (same as then clause
with “ right”and “left” exchanged
16 color [root [T]] ←BLACK
z
1
11
14
157
5
4
2
8
z
y
Case 1 : z’s uncle y is red
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11
14
15
5
4
2
8
y
Case 2
(a)
(b)
7
Case 2
1
11
14
155
4
28
z y
Case 3
1
7
14
15
5
4
2z
7
11
8
(c)
(d)
Case 1-1: z’s uncle y is red
α
α
C
D
B z
yA
β γ
δ ε
C
D
B z
A
β γ
δ ε
new z
α
C
DBz
y
A
β
γ δ ε
C
DBz
y
A
βα
γ δ ε
new z
做法 : 改變 parent, uncle, grandparent 的 color
Problem:連續雙 red α β γ δ ε :Black height 不變
Case 1-2: z’s uncle y is red
α
C
D
B z
yA
β
γδ
ε ε
α
C
DB
z
y
Aβγ
δ ε
new z
做法 : 改變 parent, uncle, grandparent 的 color
Problem:連續雙 redα β γ δ ε :Black height 不變If c’s parent is red? Continue…
C
D
B z
yA
β
γδ
new z
α
C
DB
z
y
Aβγ
δ ε
Case 2-1: z’s uncle y is black and z is a right child
Case 3-1: z’s uncle y is black and z is a left child
α
C
B z
yA
β γ
δ
case 3-1case 2-1
C
B
zy
A
βα
γ
δ
B
Az
βα γ δ
C
Left rotation
Change color: parent & grandparight rotation
α
C
B z
yA
β γ
δ
Case 3-2Case 2-2
B
C z
βα γ δ
Aα
Cy
B
zA zβ
γ δ
Right rotate
4
(a)
(b)
(c) Case 3-2
(d)
4
7
4
7
12
4
7
12 4
7
12
15
(e) Case 1-2
+root must be black
insertion
(g)
4
7
12
153
(f)
4
7
12
15
5
(h) Insert 5
3 15
4 12
7
5
(i)Insert 14
Case 2-2
3 15
4 12
7
14
5
(j)
3 1512
4 14
7
5
(k)
3 1512
4 14
7
18
5
(l)
3 1512
4 14
7
18
5
(m) Case 2-2
3 1512
4 14
7
18
16
5
(n)
3 1612
4 14
7
1815
5
(o) Insert 17
Case 1-2
3 1612
4 14
7
1815
17
5
(p) Case 3-2
3 1612
4 14
7
1815
17
12
(q)
1815
7 16
14
173
4
5
Insertion complexity
• The insertion of a key-element item in a red-black tree storing n items can be done in O(log n) time and at most O(log n) recolorings and one trinode restructuring (a restructure operation).
RB-DELETE(T,z)
1 if left[z]=nil[z] or right[z]=nil[T]
2 then y ←z
3 else z ←TREE-SUCCESSOR(z)
4 if left[y] ≠ nil[T]
5 then x← left[y]
6 else x ← right[y]
7 p[x] ← p [y]
8 if p[y]= nil[T]
9 then root [T] ← x
10 else if y=left [p[z]]
11 then left [p[z]] ← x
12 else right [p[z]] ← x
13 if y ≠z
14 then key [z] ← key [y]
15 copy y’s satellite data into z
16 if color [y] = BLACK
17 then RB-DELETE-FIXUP(T,x)
18 return y
RB-DELETE FIXUP(T,x) //y 為真正被 deleted 之 node, x 是 y 的 right or left child1 While x ≠ root[T] and color[x] =BLACK
2 do if x =left [p[x]]
3 then w ← right [p[x]]
4 if color[w] = RED
5 then color[w] ← BLACK //Case 1
6 color [p[x]] ← RED //Case 1
7 LEFT-ROTATE(T,p[x]) //Case 1
8 w ← right [p[x]] //Case 1
9 if color [left[w]] = BLACK and color [right[w]]= BLACK
10 then color[w] ← RED //Case 2
11 x ← p[x] //Case 2
12 else if color [left[w]] = BLACK
13 then color [left[w]] ← BLACK //Case 3
14 color[w] ← RED //Case 3
15 RIGHT-ROTATE(T,w) //Case 3
16 w ←right [p[x]] //Case 3
17 color[w] ← color [p[x]] //Case 4
18 color [p[x]] ← BLACK //Case 4
19 color [right[w]] ← BLACK //Case 4
20 LEFT-ROTATE(T,p[x]) //Case 4
21 x ← root[T] //Case 4
22 else (same as then clause with”right”and”left”exchanged)
23 color[x] ← BLACK // 若 x is red, 改為 black, black height 即能維持
B
Ax
βα
Case 1
δγ ζε
D
C E
w
D
x
E wB
A C
βα δγζεnew w
(a)
B
Ax
βα
Case 2
δγ ζε
D
C E
w
new x(b) c B
A
βα
δγ ζε
D
C E
c
:Red or black
(C)
(d)
y 為真正被 deleted 之 node, x 是 y 的 right or left child
restructure
Reduce 1 black height
recolor
B
Ax
βα
Case 3
δγ ζε
D
C E
w
(c)
B
Ax
βα
Case 4
δγ ζε
D
C E
w
(d) c
cnew w
B
A
βαδ
γ
ζε
C
D
E
c
x
c’
D
EB
A C
βα δγζε
c’
c
new x=root[T]
Case 1: x’s sibling w is red
Case 2: x’s sibling w is black, and both of w’s children are black
Case 3: x’s sibling w is black, w’s left child is red, and w’s right child is black
Case 4: x’s sibling w is black, and w’s right child is red
•If v is a 2-node, then keep the (black) children of v as is.
•If v is a 3-node, then create a new red node w, give v’s first two (black) children to w, and make w and v;s third child be the two children of v.
•If v is a 4-node, then create two new red nodes w and z, give v’s first two (black) children to w, give v’s last two (black) children to z, and make w and z be the two children of v.
12
(a) initial
1815
7 16
14
173
4
5
deletion
12
(b) Delete 3
1815
7 16
14
17
4
5
(c) Delete 12
1815
7 16
14
17
4
5
restructure
7
(d)
1815
5 16
14
17
4 Delete 17
7
(e)
1815
5 16
14
4 Delete 18
7
(f)
15
5 16
14
4
7
(g) After recoloring
15
5 16
14
4Delete 15
7
(h)
5 16
14
4
Delete 16
7
(i)
5
14
47
(j)
5
4 14
7
(k)
5
4 14
adjustment
recoloring
15
13 14
6 7 8
15
13
14
(a)
(b)
(c)
14
13
or
7
86
Insertion:
Case 1: The Sibling w of v is Black.
30
20
10
10
20
30
u
v
z w
u
v
zw
30
20
10
u
vz w
10
20
30
u
vzw
(a)
black
Double red
20b
a c
(b) After a trinode restructuring
3010
•Take node z, its parent v, and grandparent u, and temporarily relabel them as a,b,and c, in left-to-right order, so that a, b, c will be visited in this order by an inorder tree traversal.•Replace the grandparent u with the node labeled b, and nodes a and c the children of b, keeping inorder relationships unchanged.
Case 2: The Sibling w of v is Red.
10 20 30 40
… 30 …
10 20 40
30u
v w
(a)
4020
10z
30u
v w
(b)
4020
10z
… …
recoloring
… 30 …
10 20
30x
y r4020
10z
40
… …
(a)
Deletion: case 1: the sibling y of r is black and Has a red child z.
… 30 …
10 20
40
… …
(b)
30
20
10
x
y
zr
40
… 20 …
10 30
40
… …
(c)After restructure
20b
a c3010
40
r
(a)(b)
10 30 …
20
30x
yr
4020
40
(a)
10
…
Case 2: the sibling y of r is black and both children of y are black.
20 30
30x
yr
4020
40
(b)
10 ……
10
After recoloring
30
20
40
30x
yr
4020
(a)
20 30
30x
yr
4020
40
(b) After recoloring
20 30
(a)
… 10 …
40
… …
30x
y r4020
10z
Case 3: the sibling y of r is red.
20 30
(b) After adjustment
… 10 …
40
… …
20y
z x3010
40
r
Deletion complexity
• The algorithm for removing an item from a red-black tree with n items takes O(log n) time and performs O(log n) recolorings and at most one adjustment plus one additional trinode restructuring. Thus it perform at most two restructure operations.
2
5
(1)Insert 2 (2)Insert 1
2
1
(3)Insert 3
2
1 3
Change Color
2
1 3
(4)Insert 7
7
2
1 3
(5)Insert 5
7
2
1 3 RL
7
2
1 5
3
(6)Insert 8
8
7
2
1 5
3
(7)Insert 9
8
7
2
1 5
3
9
RR5
2 8
9731
(8)Insert 11
5
2 8
9731
11
(9)Insert 10
5
2 8
9731
11
10
RL
5
2 8
10731
119
(1)Delete 7
5
2 8
10731
119
Rotate
5
2 10
11831
97
Del 7
5
2 10
11831
9
(2)Delete 9
5
2 10
11831
9
Del 9
5
2 10
11831
最主要的三個判斷判斷一：爸爸是爺爺的左 or 右兒子判斷二： uncle 是不是紅的判斷三：我是爸爸的左 or 右兒子
所以會有八種可能 請往下看( 注意：例圖不是由樹根畫起
而是最低的三層節點。 )
爸爸是爺爺的左兒子系列Case 1
Uncle 是紅的
Case 2
在爸爸右邊
Case 3
在爸爸左邊
爸爸是爺爺的右兒子系列Case 1
Uncle 是紅的
Case 2
在爸爸右邊
Case 3
在爸爸左邊
8
8 加入 5
5
8 加入 7
7
8
5
8
7
5
1 變色
Case 3
2 右轉 7
5
5
8
7
Case 2
左轉
加入 8
8
7
5
Case 1
8
7
5
4 4
5 8
1 變色
2 根黑色7
4
58
7
6
4
5 8
7
6
2
Case 1 8
7
2
4 6
5
變色
加入 4
加入 2 加入 6
8
7
2
4 6
5 8
7
2
4 6
5
3
加入 3
1
8
7
2
4 6
5
3
Case 1
1
8
7
2
4 6
5
3
變色
Case 3
1
8
7
2
4 6
5
3
1 變色2 右轉
1
82
4
6
5
3
7
加入 1
exercise
• Complete the following algorithm: RB-DELETE FIXUP(T,x) & RB-INSERT-FIXUP(T,z)
• 實作習題四 : 以 java 實作 red-black tree (similar to 習題 1)