Advanced Trees Part III Briana B. Morrison Adapted from Alan Eugenio & William J. Collins

Advanced TreesPart III

Briana B. Morrison

Adapted from Alan Eugenio &

William J. Collins

Advanced Trees (Part III) 2

Topics

2-3-4 Search Trees Red Black Trees


2-3-4 Trees

2-3-4 Trees are just like 2-3 trees, except on inserts and deletes, the split (and/or merge) takes place before the insert (it’s done on the way down the tree rather than on the way back up).

2-3-4 trees are implemented, as a red-black tree.


3 Possible Kinds of Nodes

A

2 -n o d e

v alu e < A v alu e > A

A B

3 -n o d e

A < B

v alu e < A A < v alu e < B v alu e > B

A B C

4 -n o d e

A < B < C

v alu e < A A < v alu e < B v alu e > CB < v alu e < C


2-3-4 Tree Example

2 1 5 3 5 5 5

1 2

2 54 8 1 0

115 7 9


Example of Insertion in a 2-3-4 Tree

A

A B C

S T VU

A

B

S T VU

C


Another Insertion Example of 2-3-4 Tree

2 1 2 1 52 2 1 5

In s ert 2 In s ert 1 2In s ert 1 5

1 52

1 2

Sp lit 4 -n o d e (2 , 1 2 , 1 5 )

1 5

1 2

2 4

In s ert 4

1 5

1 2

2 4 8

Insert: 2, 15, 12, 4, 8, 10, 25, 35, 55, ….11

2

4 1 2

1 58 2

4 1 2

1 58 1 0

Sp lit 4 -n o d e (2 , 4 , 8 ) In s ert 1 0

2

4 1 2

8 1 0 1 5 2 5 2

4 1 2

2

4 1 2 2 5

8 1 0 1 5 3 5 2

4 1 2 2 5

8 1 0 1 5 3 5 5 5

1 5 2 5 3 58 1 0

I n s er t 2 5 I n s er t 3 5

I n s er t 5 5S p lit 4 - n o d e ( 1 5 , 2 5 , 3 5 )


Example (Cont…)

2 8 1 0 1 5 3 5 5 5

2 54

1 2

2 8 1 0 1 1 1 5 3 5 5 5

2 54

1 2

Sp lit 4 -n o d e (1 5 , 2 5 , 3 5 ) In s ert 1 1

2 1 5 3 5 5 5

2 5

1 2

4 1 0

1 18 2 1 5 3 5 5 5

2 5

1 2

4 1 0

1 18 9

Sp lit 4 -n o d e (8 , 1 0 , 1 1 ) In s ert 9

(4, 12, 25)

Insert 11, 9, …5


Example (Cont…)

2 1 5 3 5 5 5

2 5

1 2

4 8 1 0

1 195 7

2 1 5 3 5 5 5

2 5

1 2

4 1 0

1 15 8 9 2 1 5 3 5 5 5

2 5

1 2

4 8 1 0

1 195

In s ert 5 Sp lit 4 -n o d e (5 , 8 , 9 )

Insert 5, 7


A RED-BLACK TREE IS ANOTHER

KIND OF BALANCED BINARY SEARCH

TREE.


A RED-BLACK TREE IS A BINARY SEARCH TREE

IN WHICH THE ROOT ITEM IS COLORED

BLACK, EVERY OTHER ITEM IS COLORED

RED OR BLACK, AND

1. (RED RULE) A RED ITEM CANNOT HAVE

ANY RED CHILDREN;

2. (PATH RULE) THE NUMBER OF BLACK

ITEMS IS THE SAME IN ANY PATH FROM

THE ROOT ITEM TO AN ITEM WITH

NO CHILDREN OR WITH ONE CHILD.


Red-Black Tree

A red-black tree can also be defined as a binary search tree that satisfies the following properties: Root Property: the root is black External Property: every leaf is black Internal Property: the children of a red node are black Depth Property: all the leaves have the same black depth

9

154

62 12

7

21


Binary Search Tree and Red/Black Tree

1

2

5

4

3

2

4

53

1

D egen erat e s earch t ree red -b lack t ree


THE FOLLOWING ARE RED-BLACK

TREES:


60

30 80

20 50 90


60

30 80

20 50 90

40


WHAT ABOUT THIS?

60

30 80

20 50 90

40

THIS IS NOT A RED-BLACK TREE: THE PATH FROM60 TO 80 HAS ONLY ONE BLACK ELEMENT!


50

30 90

20 40 80 131

60 85 100 150

140 160

180

THIS RED-BLACK TREE IS NOT AN AVL TREE. WHY?


Red Black Trees as B-Trees of Order 4 Start with a B-tree of order 4 (so each node contains

1, 2, or 3 entries) Convert a node with 3 entries into a binary search

tree by:


Red-Black Trees

B

CA

A B C

S T U VS T

R ep res en t at io n w it h a b lackp aren t an d t w o red ch ild ren

4 -n o d e (A , B , C )in a 2 -3 -4 T ree

U V

A B

ST

U

3 -n o d e (A , B )in a 2 -3 -4 T ree

A

BS

R ep res en t at io n w it h a b lackp aren t an d a red righ t ch ild

T U

B

A

S T

R ep res en t at io n w it h a b lackp aren t an d a red left ch ild

U


From (2,4) to Red-Black Trees A red-black tree is a representation of a (2,4) tree by means of a

binary tree whose nodes are colored red or black In comparison with its associated (2,4) tree, a red-black tree has

same logarithmic time performance simpler implementation with a single node type

2 6 73 54

4 6

2 7

5

3

3

5OR


Converting a 2-3-4 Tree to Red-Black Tree Example

1 2 1 5

8 1 0 2 0

9 3 0 4 01 3 4 1 2 1 59 3 0 4 01 3 4

1 0

8 2 0

1 2 1 5 3 0 4 0

1 0

8 2 0

3 9

1 4

1 0

8 2 0

3 9

1 4 3 0

1 2 4 0

1 5


CLAIM: THE HEIGHT OF ANY RED-

BLACK TREE IS LOGARITHMIC IN n.


WHAT IS THE MINIMUM HEIGHT OF A

RED-BLACK TREE?

75

40 90

20 60 80 100

10 30 50


SUPPOSE A RED-BLACK TREE IS

COMPLETE, WITH ALL BLACK

ELEMENTS, EXCEPT FOR RED

LEAVES AT THE LOWEST LEVEL.

THEN THE HEIGHT OF THAT TREE IS,

APPROXIMATELY, log2n.


WHAT IS THE MAXIMUM HEIGHT? 50 30 90 20 40 80 131 60 85 100 150 140 160 185


SUPPOSE A RED-BLACK TREE HAS ALL BLACK

ELEMENTS, EXCEPT THAT ONE PATH FROM THE

ROOT TO A LEAF HAS AS MANY RED ELEMENTS AS

POSSIBLE. THEN THE LENGTH OF THAT PATH, AND

THE HEIGHT OF THE TREE, WILL BE MAXIMAL.

THE LENGTH OF THAT PATH IS APPROXIMATELY

TWICE THE MINIMAL HEIGHT, SO THE MAXIMUM

HEIGHT OF A RED-BLACK TREE WILL BE,

ROUGHLY, 2log2 (n).


THERE ARE FOUR ASSOCIATIVECONTAINER CLASSES IN THESTANDARD TEMPLATE LIBRARY:

set: ITEM = KEY; NO DUPLICATES ALLOWED

multiset: ITEM = KEY; DUPLICATES ALLOWED

map: KEY IS PART OF ITEM; NO DUPLICATES ALLOWED

multimap: KEY IS PART OF ITEM;DUPLICATES ALLOWED


Red-Black Tree Algorithms

Find – just like binary search tree Performance?

Insertion Deletion


THERE ARE FIVE STEPS TO ACCOMPLISH THE INSERTION OF THE ITEM v INTO AN rb_tree OBJECT: 1. CREATE A NODE POINTED TO BY x;

2. SET value (x) TO v;

3. INSERT *x AS A LEAF (WITH A BinSearchTree-STYLE INSERTION), AND SET color (x) TO red;

4. FIX UP THE TREE BY RE-COLORING AND RESTRUCTURING, IF THE RED RULE IS VIOLATED (PATH RULE OK: RED INSERTION);

5. SET color (root( )) TO black.


60

30 80

20 50 70 90

ADD 40?


60 30 80 20 50 70 90

40

VIOLATES RED RULE,NEED TO FLIP SOME COLORS (50)


60 30 80 20 50 70 90

40

VIOLATES PATH RULE, NEED TO FLIP SOME COLORS (30)


60 30 80 20 50 70 90

40

VIOLATES RED RULE,NEED TO FLIP SOME COLORS (20)


60 30 80 20 50 70 90

40

ADD 35?


60 30 80 20 50 70 90

40 35

THIS TREE VIOLATES THE RED RULE.


60 30 80 20 50 70 90

40 35

THIS TREE VIOLATES THE PATH RULE.


ROTATION TO THE RESCUE!


60

30 80

20 40 70 90

35 50


WHEN WILL STEP 4 BE NEEDED?

NOT IF (THE NODE POINTED TO BY) x

IS THE ROOT, BECAUSE THEN STEP 5

WILL SET x’S COLOR TO black.

ALSO, IF color (parent (x)) IS black, THEN

NO RESTRUCTURING IS NEEDED

BECAUSE THE RED RULE IS NOT

VIOLATED.


STEP 4 IS ACCOMPLISHED IN A LOOP:

while (x != root() && color(parent(x)) == red){

} // while

color (root( )) = black; // STEP 5


THE LOOP CONSISTS OF THREE CASES

WHEN parent (x) IS A LEFT CHILD, AND

A SYMMETRIC THREE CASES WHEN

parent (x) IS A RIGHT CHILD. IN ALL

CASES, THE ACTION TAKEN DEPENDS

ON THE SIBLING OF parent (x).


ASSUME THAT parent (x) IS A LEFT

CHILD. LET y BE (A POINTER TO) THE

SIBLING OF parent (x). CASE 1: color (y) is red FOR EXAMPLE, SUPPOSE 10 HAS JUST

BEEN INSERTED INTO THE

FOLLOWING TREE:


100

40 120

30 80 135

10 y ACTION: color(parent(x)) = black; x color(y) = black; color(parent(parent(x))) = red; x = parent(parent(x));


100

x 40 120

30 80 135

10 y

THE LOOP NOW TERMINATES BECAUSE

color (parent (x)) IS black. AND THIS IS A RED-BLACK

TREE.


CASE 2: color (y) IS black, AND x IS A RIGHT CHILD RECALL THAT WHEN y IS NULL, y

BECOMES A POINTER TO NIL, WHOSE

COLOR IS black.

FOR EXAMPLE, SUPPOSE 20 HAS JUST

BEEN INSERTED INTO THE

FOLLOWING TREE:


100

40 120

30 80 135

10 y IS NULL ACTION: x = parent (x); 20 x rotate_left (x);


100 40 120 30 80 135 20 y is NULL 10 x

THIS LEADS US TO CASE 3.


CASE 3: color (y) IS black, AND x IS A

LEFT CHILD

FOR EXAMPLE, CASE 3 APPLIES TO

THE TREE IN THE PREVIOUS SLIDE:


100

40 120

30 80 135

20 y is NULL

10 x


100 40 120 30 80 135 20 y is NULL 10 x ACTION:

color (parent (x)) = black; color (parent (parent (x))) = red; rotate_right (parent (parent (x)));


100 40 120 20 80 135 10 30 x

THE PARENT OF x IS NOW black, SO WE ARE DONE.


ADDITIONAL LOOP ITERATIONS

OCCUR ONLY WHEN CASE 1 APPLIES,

AND THEN x IS MOVED TWO LEVELS

UP THE TREE. SO FOR RECOLORING

AND RESTRUCTURING, worstTime(n) IS

LOGARITHMIC IN n. (RECALL THAT

THE HEIGHT OF A RED-BLACK TREE

IS ALWAYS LOGARITHMIC IN n.)


FOR THE ORIGINAL INSERTION,

BinSearchTree-STYLE, worstTime(n) AND

averageTime(n) ARE LOGARITHMIC IN n

FOR THE SAME REASON (THE HEIGHT

OF THE TREE IS LOGARITHMIC IN n).

SO FOR THE insert METHOD,

worstTime(n) AND averageTime(n) ARE

LOGARITHMIC IN n.


Building A Red-Black Tree

Let’s build a red-black tree both ways As a 2-3-4 tree As a red-black tree

Insert the following values 2, 15, 12, 4, 8, 10, 25, 35


Building A Red-Black Tree2 2

In s ert 2m ak e ro o tB L A C K

R ed -B lack T ree 2 - 3 - 4 T r e e

22

R e d- B la c k T r e e

B L A C K

2 1 2 1 51 5

22 -3 -4 T ree

R ed -B lack T ree

1 2

A B

C

DX

G

1 52

1 2

C A B D

D o u b le l eftro tat io n

P

2 4

2 -3 -4 T reeR ed -B lack T ree

1 2

1 51 52

1 2

41 52

1 2

1 52

1 2

C o lo r F l ip

1 52

1 2

M ak e ro o tB L A C K

In s ert 4

4

Insert 15 and 12

Insert 4


2 4 8

1 54


1 21 2

1 582

1 54

2 -3 -4 T ree

R ed -B lack T ree

1 2

1 5

82

4 1 2

2 8 1 0

1 54

1 2

82

1 54

1 2

82

1 0

C o lo r fl ip In s ert 1 0

8

1 5

1 2

82

1 2

82

2 5


4 1 2

2 8 1 0

4

1 0

1 5 2 52 5

In s ert 2 5


4 1 2

2 8 1 01 5

4

2

1 0

1 5 2 5 3 53 5

In s ert 3 5

Insert 8

Insert 10

Insert 25 Insert 35


Insertion

To perform operation insertItem(k, o), we execute the insertion algorithm for binary search trees and color red the newly inserted node z unless it is the root We preserve the root, external, and depth properties If the parent v of z is black, we also preserve the internal property and

we are done Else (v is red ) we have a double red (i.e., a violation of the internal

property), which requires a reorganization of the tree Example where the insertion of 4 causes a double red:

6

3 8

6

3 8

4z

v v

z


Remedying a Double Red

Consider a double red with child z and parent v, and let w be the sibling of v

4

6

7z

vw2

4 6 7

.. 2 ..

Case 1: w is black The double red is an incorrect

replacement of a 4-node Restructuring: we change the 4-

node replacement

Case 2: w is red The double red corresponds

to an overflow Recoloring: we perform the

equivalent of a split

4

6

7z

v

2 4 6 7

2w


Restructuring

A restructuring remedies a child-parent double red when the parent red node has a black sibling

It is equivalent to restoring the correct replacement of a 4-node The internal property is restored and the other properties are

preserved

4

6

7z

vw2

4 6 7

.. 2 ..

4

6

7

z

v

w2

4 6 7

.. 2 ..


Restructuring (cont.)

There are four restructuring configurations depending on whether the double red nodes are left or right children

2

4

6

6

2

4

6

4

2

2

6

4

2 6

4


Recoloring

A recoloring remedies a child-parent double red when the parent red node has a red sibling

The parent v and its sibling w become black and the grandparent u becomes red, unless it is the root

It is equivalent to performing a split on a 5-node The double red violation may propagate to the grandparent u

4

6

7z

v

2 4 6 7

2w

4

6

7z

v

6 7

2w

… 4 …

2


Analysis of Insertion

Recall that a red-black tree has O(log n) height

Step 1 takes O(log n) time because we visit O(log n) nodes

Step 2 takes O(1) time Step 3 takes O(log n) time

because we perform O(log n) recolorings, each

taking O(1) time, and at most one restructuring taking

O(1) time Thus, an insertion in a red-

black tree takes O(log n) time

Algorithm insertItem(k, o)

1. We search for key k to locate the insertion node z

2. We add the new item (k, o) at node z and color z red

3. while doubleRed(z)if isBlack(sibling(parent(z)))

z restructure(z)return

else { sibling(parent(z) is red } z recolor(z)


EXERCISE: INSERT 85 INTO THE FOLLOWING RED-

BLACK TREE: 100 40 120 20 80 135 10 30


EXERCISE: INSERT 15 INTO THE FOLLOWING RED-

BLACK TREE, AND FIX AFTER INSERTION: 100 40 120 20 80 135 10 30


Deletion

To perform operation remove(k), we first execute the deletion algorithm for binary search trees

Let v be the internal node removed, w the external node removed, and r the sibling of w If either v of r was red, we color r black and we are done Else (v and r were both black) we color r double black, which is a

violation of the internal property requiring a reorganization of the tree Example where the deletion of 8 causes a double black:

6

3 8

4

v

r w

6

3

4

r


Remedying a Double Black

The algorithm for remedying a double black node w with sibling y considers three casesCase 1: y is black and has a red child We perform a restructuring, equivalent to a transfer , and we are done

Case 2: y is black and its children are both black We perform a recoloring, equivalent to a fusion, which may propagate

up the double black violation

Case 3: y is red We perform an adjustment, equivalent to choosing a different

representation of a 3-node, after which either Case 1 or Case 2 applies Deletion in a red-black tree takes O(log n) time


Red-Black Tree Reorganization

Insertion remedy double red

Red-black tree action (2,4) tree action result

restructuringchange of 4-node representation

double red removed

recoloring splitdouble red removed or propagated up

Deletionremedy double black

Red-black tree action (2,4) tree action result

restructuring transfer double black removed

recoloring fusiondouble black removed or propagated up

adjustmentchange of 3-node representation

restructuring or recoloring follows


Tree Summary

General Trees Multiway search trees B trees 2-3 search trees Tries AVL Trees 2-3-4 trees Red-black trees


BST, Red-Black and AVL Example

B in ary Search T ree (a)

5 0

7 8

7 5

8 0

7 0

9 0

6 0

R ed -B lack T ree (b )

7 8

8 07 05 0

9 06 0

7 5

8 07 5

7 85 0

9 06 0

7 0

A VL T ree (c)

1 0 01 0 0

1 0 0

Documents

Advanced Trees Part III Briana B. Morrison Adapted from Alan Eugenio & William J. Collins