deﬁnisali su ih gibs i sedžvik 1978. godine; crveno-crna

OvervievStražar u crveno-crnom stabluUbacivanje cvoraBrisanje cvora

Crveno-crna stabla

– Definisali su ih Gibs i Sedžvik 1978. godine;

Crveno-crna stabla su binarna stabla, kod kojih je zaskladištenje svakog cvora potreban još jedan bit:

njegova boja koja može biti crvena ili crna.

Vodeci racuna o boji cvorova na putu od korena do bilo koglista, u crveno-crnom stablu mora biti ispunjen uslov da je tajput najviše dva puta duži od bilo kog drugog.

To znaci da su ova stabla na neki nacin izbalansirana.

Svaki cvor ima pet atributa:

– color, key, left, right, p.


Crveno-crna stabla (cont.)

Ako neki cvor nema roditelja ili dete pokazivac na njih je NIL.

Pretpostavicemo da NIL pokazivaci ukazuju i na listove stabla,dok je svaki cvor koji ima neki sadržaj unutrašnji cvor stabla.

Crveno-crno stablo se karakteriše sledecim osobinama:



Svaki cvor u stablu je ili crven ili crn;

Koren je crn;

Svaki list (NIL) je crn;

Akoje cvor crven oba deteta su mu crna;

Putevi od svakog lista do korena imaju isti broj crnihcvorova;

Crveno-crna stabla se pretražuju od korena do mesta na kometreba da se nalazi element koji se ubacuje ili izbacuje, a zatimse vraca prema korenu i uspostavlja se crveno-crna strukturaukoliko je narušena.



Teorema:Visina crveno-crnog stabla koje ima n cvorova jeste najviše2 log2 (n + 1).

Dokaz cemo izvesti indukcijom po visini cvora x. Odredicemonajpre crnu visinu stabla sa korenom u x, koju cemo oznaciti sabh(x). To je broj crnih cvorova od korena stabla do lista, neracunajuci koren. Indukcijom se lako dokazuje da stablo crnevisine bh(x), sa korenom u proizvoljnom cvoru x ima najmanje2bh(x)

− 1 cvorova.

Ako je bh(x) = 0, jasno je da je x list (T.NIL) i podstablo sakorenom u x sadrži 2bh(x)

− 1 = 20− 1 = 1 − 1 = 0 cvorova,

ako izuzmemo x.



Za induktivni korak pretpostavimo da je x unutrašnji cvor stablasa dvoje dece. Svako dete ima crnu visinu bh(x) ili bh(x) − 1,zavisno od toga dali je crveno ili crno, redom. Kako je visinasvakog deteta od x manja od visine cvora x, možemo primenitiindukcijsku pretpostavku, pa svako dete odredjuje podstablo sanajmanje 2bh(x)−1

− 1 cvorova.

Dakle, podstablo sa korenom u x ima najmanje2(2bh(x)−1

− 1) + 1 = 2bh(x)− 1 cvorova.

Da bi upotpunili dokaz, pretpostavimo da stablo sa korenom xima visinu h. Prema osobinama crveno-crnog stabla barpolovina cvorova na putu od korena do lista je crna, pa jebh(x) >

h2 . Tada je broj cvorova u stablu

n > 2h2 −1

⇔ h 6 2 log2 (n + 1)


Stražar (sentinel) u crveno-crnom stablu

Da bi pojednostavili granicne uslove u crveno-crnom stablumožemo koristiti jednog stražara koji predstavlja NIL.

U crveno-crnom stablu stražar T.NIL je objekat sa istimatributima kao obican cvor u stablu. Njegova boja je crna, dokostali atributi- p, left, right i key- mogu uzeti proizvoljnevrednosti.


Stražar u crveno-crnom stablu (cont.)

Svi pokazivaci na listove, zamenjuju se pokazivacem nasentinel T.NIL.

Kao što se vidi, kada koristimo stražara, NIL dete cvora xmožemo smatrati obicnim detetom ciji je roditelj x. T.NIL uovom slucaju se koristi i kao roditelj korena stabla.

U opštem slucaju, u radu sa crveno-crnim stablima, nama suvažniji unutrašnji cvorovi, jer oni nose vrednosti kljuceva. Zatolistove možemo izostaviti pri crtanju ovih stabala.


Ubacivanje cvora u crveno-crno stablo

Procedura ubacivanja cvora u crveno-crno stablo sa n cvorovazahteva O(log2 n) vremena.

Procedura ubacivanja cvora u crveno-crno stablo je malomodifikovana procedura ubacivanja elementa z u uredjenobinarno stablo, posle cega se cvor z boji crveno.

Kako se, pri tom, može narušiti crveno-crna struktura,neophodno je primeniti levu ili desnu rotaciju za preuredjenjecvorova.


Ubacivanje cvora u crveno-crno stablo (cont.)



U opštem slucaju se crveno-crna struktura može narušiti samoako je z koren stabla ili u koliko je roditeljski cvor od z crven.



Ako je cvor z.p crven, onda on nije koren stabla, pa sigurnopostoji z.p.p.

Trebalo bi da posmatramo šest slucajeva u WHILE petlji, ali trisu simetricna, zavisno od toga da li je roditeljski cvor z.p levo ilidesno dete praroditelja z.p.p od z (zato se posmatra samojedan slucaj, kada je z.p levo dete).

Zatim se proverava ujak cvora z ciju vrednost pamti promenljivay. Ako je y crven imamo Case 1, dok u protivnom algoritamprelazi na Case 2 i Case 3. Jasno da je u sva tri slucaja cvorz.p.p crn.

Na slici je data situacija kada su i z.p i y crveni. Prebojavamo ihu crno, dok z.p.p bojimo u crveno i pomeramo y do z.p.p dvakoraka ponavljajuci tako WHILE petlju.

Case 2. Ujak od z je crn, a z je desno dete roditelja;

Case 3. Ujak od z je crn, a z je levo dete roditelja;


Brisanje cvora iz crveno-crnog stabla

Kao i druge osnovne operacije u crveno-crnom stablu sa ncvorova, brisanje cvora zahteva Olog2 n vremena.

Brisanje cvora je malo složenija operacija od ubacivanja cvora.

Procedura za brisanje cvora iz crveno-crnog stabla zasniva sena proceduri brisanja cvora iz binarnog stabla.

Najpre cemo napisati proceduru TRANSPLANT:


Brisanje cvora iz crveno-crnog stabla (cont.)

3

Red-Black Trees

• Definition: A red-black tree is a binary search tree where:– Every node is either red or black.– Each NULL pointer is considered to be a black node– If a node is red, then both of its children are black.– Every path from a node to a leaf contains the same

number of black nodes.

• Definition: The black-height of a node, n, in a red-black tree is the number of black nodes on any path to a leaf, not counting n.

4

A valid Red-Black Tree

Black-Height = 2

7

Theorem 1 – Any red-black tree with root x, has at least n = 2bh(x) – 1 internal nodes, where bh(x) is the black height of node x.

Proof: by induction on height of x.

8

Theorem 2 – In a red-black tree, at least half the nodes on any path from the root to a leaf must be black.

Proof – If there is a red node on the path, there must be a corresponding black node.

9

Theorem 3 – In a red-black tree, no path from any node, N, to a leaf is more than twice as long as any other path from N to any other leaf.

Proof: By definition, every path from a node to any leaf contains the same number of black nodes. By Theorem 2, a least ½ the nodes on any such path are black. Therefore, there can no more than twice as many nodes on any path from N to a leaf as on any other path. Therefore the length of every path is no more than twice as long as any other path

10

Theorem 4 –A red-black tree with n internal nodes has height h <= 2 lg(n + 1).

Proof: Let h be the height of the red-black tree with root x. By Theorem 2,

bh(x) >= h/2From Theorem 1, n >= 2bh(x) - 1Therefore n >= 2 h/2 – 1

n + 1 >= 2h/2

lg(n + 1) >= h/22lg(n + 1) >= h

11

Bottom –Up Insertion

• Insert node as usual in BST• Color the Node RED• What Red-Black property may be violated?

– Every node is Red or Black– Leaf nodes are Black NULLS– If node is Red, both children must be Black– Every path from node to descendant leaf must

contain the same number of Blacks

12

Bottom Up Insertion• Insert node; Color it RED; X is pointer to it• Cases

0: X is the root -- color it black1: Both parent and uncle are red -- color parent and uncle

black, color grandparent red, point X to grandparent, check new situation

2 (zig-zag): Parent is red, but uncle is black. X and its parent are opposite type children -- color grandparent red, color X black, rotate left on parent, rotate right on grandparent

3 (zig-zig): Parent is red, but uncle is black. X and its parent are both left or both right children -- color parent black, color grandparent red, rotate right on grandparent

13

X

P

G

U

P

G

U

Case 1 – U is Red

Just Recolor and move up

X

14

X

P

G

U

S X

P G

SU

Case 2 – Zig-Zag

Double Rotate X around P; X around G

Recolor G and X

15

X

P

G

U

S P

X G

S U

Case 3 – Zig-Zig

Single Rotate P around G

Recolor P and G

16

11

14

15

2

1 7

5 8

Black node Red node

Insert 4 into this R-B Tree

17

Insertion Practice

Insert the values 2, 1, 4, 5, 9, 3, 6, 7 into an initially empty Red-Black Tree

18

Asymptotic Cost of Insertion

• O(lg n) to descend to insertion point

• O(1) to do insertion

• O(lg n) to ascend and readjust == worst case only for case 1

• Total: O(log n)

Red-Black Trees

Bottom-Up Deletion

Recall “ordinary” BST Delete

• If vertex to be deleted is a leaf, just delete it.

• If vertex to be deleted has just one child, replace it with that child

• If vertex to be deleted has two children, replace the value of by it’s in-order predecessor’s value then delete the in-order predecessor (a recursive step)

Bottom-Up Deletion

1. Do ordinary BST deletion. Eventually a “case 1” or “case 2“ will be done (leaf or just one child). If deleted node, U, is a leaf, think of deletion as replacing with the NULL pointer, V. If U had one child, V, think of deletion as replacing U with V.

2. What can go wrong??

Which RB Property may be violated after deletion?

1. If U is red?

Not a problem – no RB properties violated

2. If U is black?

If U is not the root, deleting it will change the black-height along some path

Fixing the problem

• Think of V as having an “extra” unit of blackness. This extra blackness must be absorbed into the tree (by a red node), or propagated up to the root and out of the tree.

• There are four cases – our examples and “rules” assume that V is a left child. There are symmetric cases for V as a right child

Terminology• The node just deleted was U

• The node that replaces it is V, which has an extra unit of blackness

• The parent of V is P

• The sibling of V is S

Black Node

Red Node

Red or Black and don’t care

Bottom-Up DeletionCase 1

• V’s sibling, S, is Red– Rotate S around P and recolor S & P

• NOT a terminal case – One of the other cases will now apply

• All other cases apply when S is Black

Case 1 DiagramP

SV+P

S

V+

Rotate

P

V+

SRecolor


• V’s sibling, S, is black and has two black children.– Recolor S to be Red– P absorbs V’s extra blackness

• If P is Red, we’re done• If P is Black, it now has extra blackness and

problem has been propagated up the tree

Case 2 diagram

P

SV+

P+

SV

Recolor and absorb

Either extra black absorbed by P or

P now has extra blackness


• S is black

• S’s RIGHT child is RED (Left child either color)– Rotate S around P– Swap colors of S and P, and color S’s Right

child Black

• This is the terminal case – we’re done

Case 3 diagramsP

SV+P

S

V

Rotate

P

S

V

Recolor


• S is Black, S’s right child is Black and S’s left child is Red– Rotate S’s left child around S– Swap color of S and S’s left child– Now in case 3

Case 4 DiagramsP

SV+

P

S

V+Rotate

P

SV+

Recolor

65

50 80

10 60 70 90

62

Perform the following deletions, in the order specifiedDelete 90, Delete 80, Delete 70

Red Black Trees

Top-Down Insertion

Review of Bottom-Up Insertion

• In B-Up insertion, “ordinary” BST insertion was used, followed by correction of the tree on the way back up to the root

• This is most easily done recursively– Insert winds up the recursion on the way down

the tree to the insertion point– Fixing the tree occurs as the recursion unwinds

Top-Down Insertion Strategy

• In T-Down insertion, the corrections are done while traversing down the tree to the insertion point.

• When the actual insertion is done, no further corrections are needed, so no need to traverse back up the tree.

• So, T-Down insertion can be done iteratively which is generally faster

Goal of T-D Insertion

• Insertion is always done as a leaf (as in ordinary BST insertion)

• Recall from the B-Up flow chart that if the uncle of a newly inserted node is black, we restore the RB tree properties by one or two local rotations and recoloring – we do not need to make changes further up the tree

Goal (2)

• Therefore, the goal of T-D insertion is to traverse from the root to the insertion point in such a way that RB properties are maintained, and at the insertion point, the uncle is Black.

• That way we may have to rotate and recolor, but not propagate back up the tree

Possible insertion configurations

X (Red or Black)

Y Z

If a new node is inserted as a child of Y or Z, there is no problem since the new node’s parent is black


X

Y Z

If new node is child of Z, no problem since Z is black.

If new node is child of Y, no problem since the new node’s uncle (Z) is black – do a few rotations and recolor…. done


X

Y Z

If new node is inserted as child of Y or Z, it’s uncle will be red and we will have to go back up the tree. This is the only case we need to avoid.

Top-Down Traversal

X

Y Z

As we traverse down the tree and encounter this case, we recolor and possible do some rotations.

There are 3 cases.

Remember the goal – to create an insertion point at which the parent of the new node is Black, or the uncle of the new node is black.

Case 1 – X’s Parent is Black

X

ZY

P

X

Z

P

Just recolor and continue down the tree

Y

Case 2

• X’s Parent is Red (so Grandparent is Black) and X and P are both left/right children– Rotate P around G– Color P black– Color G red

• Note that X’s uncle, U, must be black because it (a) was initially black, or (b) would have been made black when we encountered G (which would have had two red children -- X’s Parent and X’s uncle)

Case 2 diagrams

X

ZY

P

G

U

SX

Z

Y

P

G

US

Rotate P around G. Recolor X, Y, Z, P and G

Case 3

• X’s Parent is Red (so Grandparent is Black) and X and P are opposite children– Rotate P around G– Color P black– Color G red

• Again note that X’s uncle, U, must be black because it (a) was initially black, or (b) would have been made black when we encountered G (which would have had two red children -- X’s Parent and X’s uncle)

Case 3 Diagrams (1 of 2)

X

ZY

P

G

U

S

X

YS

P

G

U

Z

Step 1 – recolor X, Y and Z. Rotate X around P.

Case 3 Diagrams (2 of 2)

X

YS

P

G

U

Z

P

YS

X

G

UZ

Step 2 – Rotate X around G. Recolor X and G

An exercise – insert F

D

T

W

ZV

L

J P

E K

Top-Down Insert Summary

P

X

Y Z

Case 1

P is Black

Just Recolor

P

X

Y Z

Case 2P is RedX & P both left/right

P

X

Y Z

G

P

X

Y Z

GRecolorX,Y,Z

P

X

Y Z

GRotate P around GRecolor P,G

Case 3P is RedX and P are opposite children

P

X

Y Z

GRecolor X,Y,Z Rotate X around P

X

P

Y Z

G

Rotate X around GRecolor X, G

Recolor X,Y,Z

X

P

Y Z

G

Red Black Trees

Top-Down Deletion

Recall the rules for BST deletion

• If vertex to be deleted is a leaf, just delete it.

• If vertex to be deleted has just one child, replace it with that child

• If vertex to be deleted has two children, replace the value of by it’s in-order predecessor’s value then delete the in-order predecessor (a recursive step)

What can go wrong?

1. If the delete node is red?

Not a problem – no RB properties violated

2. If the deleted node is black?

If the node is not the root, deleting it will change the black-height along some path

The goal of T-D Deletion

• To delete a red leaf

• How do we ensure that’s what happens?– As we traverse the tree looking for the leaf to

delete, we change every node we encounter to red.

– If this causes a violation of the RB properties, we fix it

Bottom-Up vs. Top-Down

• Bottom-Up is recursive– BST deletion going down the tree (winding up

the recursion)– Fixing the RB properties coming back up the

tree (unwinding the recursion)

• Top-Down is iterative– Restructure the tree on the way down so we

don’t have to go back up

Terminology

• Matching Weiss text section 12.2– X is the node being examined– T is X’s sibling– P is X’s (and T’s) parent– R is T’s right child– L is T’s left child

• This discussion assumes X is the left child of P. As usual, there are left-right symmetric cases.

Basic Strategy

• As we traverse the tree, we change every node we visit, X, to Red.

• When we change X to Red, we know– P is also Red (we just came from there)– T is black (since P is Red, it’s children are

Black)

Step 1 – Examine the root

1. If both of the root’s children are Blacka. Make the root Red

b. Move X to the appropriate child of the root

c. Proceed to step 2

2. Otherwise designate the root as X and proceed to step 2B.

Step 2 – the main caseAs we traverse down the tree, we continually

encounter this situation until we reach the node to be deleted

X is Black, P is Red, T is Black

We are going to color X Red, then recolor other nodes and possibly do rotation(s) based on the color of X’s and T’s children

2A. X has 2 Black children2B. X has at least one Red child

P

TX

Case 2AX has two Black Children

2A1. T has 2 Black Children

2A2. T’s left child is Red

2A3. T’s right child is Red

** if both of T’s children are Red, we can do either 2A2 or 2A3

Case 2A1X and T have 2 Black Children

P

TX

P

TX

Just recolor X, P and T and move down the tree

Case 2A2

P

TX

L

X has 2 Black Children and T’s Left Child is Red

Rotate L around T, then L around PRecolor X and P then continue down the tree

L1 L2

P T

X

L

L1 L2

Case 2A3

P

TX

X has 2 Black Children and T’s Right Child is Red

Rotate T around PRecolor X, P, T and R then continue down the tree

R1 R2

P R

X

T

R2R1R

L L

Case 2BX has at least one Red child

Continue down the tree to the next level

If the new X is Red, continue down again

If the new X is Black (T is Red, P is Black)

Rotate T around P

Recolor P and T

Back to main case – step 2

Case 2B Diagram

P

X T

Move down the tree.

P

X T

P

T X

If move to Black child (2B2)Rotate T around P; Recolor P and TBack to step 2, the main case

If move to the Red child (2B1) Move down again

Step 3

Eventually, find the node to be deleted – a leaf or a node with one non-null child that is a leaf.

Delete the appropriate node as a Red leaf

Step 4Color the Root Black

Example 1Delete 10 from this RB Tree

15

17

1620

23181310

7

12

6

3

Step 1 – Root has 2 Black children. Color Root Red

Descend the tree, moving X to 6

Example 1 (cont’d)

15

17

1620

23181310

7

12

6

3

One of X’s children is Red (case 2B). Descend down the tree, arriving at 12. Since the new X (12) is also Red (2B1), continue down the tree, arriving at 10.

X


15

17

1620

23181310

7

12

6

3

Step 3 -Since 10 is the node to be deleted, replace it’s value with the value of it’s only child (7) and delete 7’s red node

X


15

17

1620

2318137

12

6

3

The final tree after 7 has replaced 10 and 7’s red node deleted and (step 4) the root has been colored Black.


15

17

1620

1310

12

6

3

42

Step 1 – the root does not have 2 Black children.

Color the root red, Set X = root and proceed to step 2


15

17

1620

1310

12

6

3

42

X

X has at least one Red child (case 2B). Proceed down the tree, arriving at 6. Since 6 is also Red (case 2B1), continue down the tree, arriving at 12.


15

17

1620

1310

12

6

3

42

X

X has 2 Black children. X’s sibling (3) also has 2 black children.Case 2A1– recolor X, P, and T and continue down the tree, arriving at 10.

P

T


15

17

1620

1310

12

6

3

42

P

X T

X is now the leaf to be deleted, but it’s Black, so back to step 2.X has 2 Black children and T has 2 Black children – case 2A1

Recolor X, P and T. Step 3 -- Now delete 10 as a red leaf.Step 4 -- Recolor the root black

Example 2 Solution

15

17

1620

13

12

6

3

42


15

1311

12

10

5

73

6 9 2 4

Valid and unaffected Right subtree

Step 1 – root has 2 Black children. Color Root red.

Set X to appropriate child of root (10)


15

1311

12

10

5

73

6 9 2 4

X

X has one Red child (case 2B)

Traverse down the tree, arriving at 12.


15

1311

12

10

5

73

6 9 4

X

Since we arrived at a black node (case 2B2) assuring T is red and P is black), rotate T around P, recolor T and P

Back to step 2

P

T

2


15

1311

12

10

5

73

6 9 4

XP

T

2

Now X is Black with Red parent and Black sibling.X and T both have 2 Black children (case 2A1)Just recolor X, P and T and continue traversal

Example 3 (cont’d)15

1311

12

10

5

73

6 9 4X

P

T 2

Having traversed down the tree, we arrive at 11, the leaf to be deleted, but it’s Black, so back to step 2.X and T both have two Black children. Recolor X, P and T.Step 3 -- delete 11 as a red leaf. Step 4 -- Recolor the root black

Example 3 Solution

13

12

10

5

73

6 9 4 2

15

deﬁnisali su ih gibs i sedžvik 1978. godine; crveno-crna

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