C7_Cay

Chng 7: CY (Tree)

Chng 7: Cy (Tree)

Ni dung

*
Cu trc cy (Tree)Cu trc cy nh phn (Binary Tree)Cu trc cy nh phn tm kim (Binary Search Tree)Cu trc cy nh phn tm kim cn bng (AVL Tree)
Chng 7: Cy (Tree)

Tree inh nghia

*
Cy l mt tp hp T cc phn t (gi l nt ca cy) trong c 1 nt c bit c gi l gc, cc nt cn li c chia thnh nhng tp ri nhau T1, T2 , ... , Tn theo quan h phn cp trong Ti cng l mt cyA tree is a set of one or more nodes T such that:i. there is a specially designated node called a rootii. The remaining nodes are partitioned into n disjointed set of nodes T1, T2,,Tn, each of which is a tree
Chng 7: Cy (Tree)

Tree V d

*
S t chc ca mt cng ty
Cng ty A

R&D

Kinh doanh

Tai vu

San xuat

TV

CD

Amplier

Noi a

Quoc te

Chau au

My

Cac nc

Chng 7: Cy (Tree)

Tree V d
Cy th mc
*

Chng 7: Cy (Tree)

Tree V d

*

Chng 7: Cy (Tree)

Tree V d

This is a tree because it is a set of nodes {A,B,C,D,E,F,G,H,I}, with node A as a root node and the

remaining nodes partitioned into three disjointed sets {B,G,H,I}, { C,E,F} and {D}, respectively. Each of

these sets is a tree because each satisfies the aforementioned definition properly.

Chng 7: Cy (Tree)

Tree V d
Khng phai cy
*

Nhn xet: Trong cu trc cy khng tn ti chu trnh

Even though this is a set of nodes {A,B,C,D,E,F,G,H,I}, with node A as a root node, this is not a tree

because the fact that node E is shared makes it impossible to partition nodes B through I into

disjointed sets.

Chng 7: Cy (Tree)

Tree - Mt s khi nim c bn
Bc ca mt nt (Degree of a Node of a Tree): L s cy con ca nt . Nu bc cua mt nut bng 0 thi nut o goi la nut la (leaf node)Bc ca mt cy (Degree of a Tree):L bc ln nht ca cc nt trong cy. Cy c bc n th gi l cy n-phnNt gc (Root node): L nt khng c nt chaNt l (Leaf node): L nt c bc bng 0
*

Chng 7: Cy (Tree)

Tree - Mt s khi nim c bn

*
Nt nhanh: L nt c bc khc 0 v khng phi l gcMc ca mt nt (Level of a Node): Mc (gc (T) ) = 0Gi T1, T2, T3, ... , Tn l cc cy con ca T0 Mc(T1) = Mc(T2) = ... = Mc(Tn) = Mc(T0) + 1
Chng 7: Cy (Tree)

Tree V d

Chng 7: Cy (Tree)

*

Owner

Manager

Chef

Waitress

Cook

Helper

Waiter

Nut gc (root node)

Tree V d

nut l (leaf node)

*

Chng 7: Cy (Tree)

Tree V d

*

Tree nodes

Tree edges

Chng 7: Cy (Tree)

Tree V d

*

Gc(root)

Nt trong

l

cha

v

con

Chng 7: Cy (Tree)

*

Owner

Jake

Manager Chef Brad Carol

Waitress Waiter Cook Helper

Joyce Chris Max Len

A Tree Has Levels

LEVEL 0

*

Chng 7: Cy (Tree)

*

Owner

Jake



Joyce Chris Max Len

Level One

LEVEL 1

*

Chng 7: Cy (Tree)

*

Owner

Jake



Joyce Chris Max Len

Level Two

LEVEL 2

*

Chng 7: Cy (Tree)

nh ngha

*

Node 0

Node 1

Node 2

Node 3

Node 4

Node 5

Node 6

l

Gc

Node 1,2,3 con ca gc

Node 4, 5 l anh em

Node 1 l cha ca

Nodes 4,5

Node 0 l t tin ca tt c cc node

Nodes 1-6 l con chu ca node 0

Chng 7: Cy (Tree)

Mt s khi nim c bn
di ng i t gc n nt x:
Px = s nhnh cn i qua k t gc n x
di ng i tng ca cy:
trong Px l di ng i t gc n X
di ng i trung bnh: PI = PT/n (n l s nt trn cy T)Rng cy: l tp hp nhiu cy trong th t cc cy l quan trng
*

Chng 7: Cy (Tree)

Depth-first Search

*

Chng 7: Cy (Tree)

Breadth-first Search

*

Chng 7: Cy (Tree)

Ni dung

*
Chng 7: Cy (Tree)

Binary Tree inh nghia

*
Cy nh phn l cy m mi nt c ti a 2 cy con
Chng 7: Cy (Tree)

Binary Tree Vi du

*

Cy con trai

Cy con phai

Hinh anh mt cy nh phn

Chng 7: Cy (Tree)

Binary Tree Vi du
Cy lch trai va cy lch phai
Chng 7: Cy (Tree)

Binary Tree Vi du
A full binary tree
Chng 7: Cy (Tree)

Binary Tree Vi du
Cy nh phn dng biu din mt biu thc ton hc:
*

Chng 7: Cy (Tree)

Binary Tree Mt s tnh cht
S nt nm mc i 2iS nt l 2h-1, vi h l chiu cao ca cyChiu cao ca cy h log2N, vi N la s nt trong cyS nt trong cy 2h-1vi h l chiu cao ca cy
*

Xem them gtrinh trang 142

A full binary tree is a binary of depth k having 2^k 1 nodes. If it has < 2^k 1, it is not a full binary tree.

Chng 7: Cy (Tree)

Binary Tree
Cy nh phn y
*

0

1

2

3

7

4

5

6

8

0

1

2

3

4

5

6

7

8

2k+1, 2k+2

k=3

y : cc node l lun nm mc cao nht v cc nt khng l nt l c y 2 nhnh con.

Chng 7: Cy (Tree)

Binary Tree - Biu din
In general, any binary tree can be represented using an array, but ?
If a binary tree is a complete binary tree, it can be represented using an array capable of holding n

elements where n is the number of nodes in a complete binary tree. If the tree is an array of n

elements, we can store the data values of the ith node of a complete binary tree with n nodes at an

index i in an array tree. That means we can

map node i to the ith index in the array, and

the parent of node i will get mapped at an index i/2, whereas

the left child of node i gets mapped at an index 2i and

the right child gets mapped at an index 2i + 1.

For example, a complete binary tree with depth k = 3,

having the number of nodes n = 5, can be represented using an array of

In general, any binary tree can be represented using an array. We see that an array representation of

a complete binary tree does not lead to the waste of any storage. But if you want to represent a binary

tree that is not a complete binary tree using an array representation, then it leads to the waste of

storage as shown in Figure 7.9.

Chng 7: Cy (Tree)


*
S dng mt bin ng lu tr cc thng tin cua mt nut:Thng tin lu tr ti nta ch nt gc ca cy con tri trong b nha ch nt gc ca cy con phi trong b nhKhai bao cu truc cy nhi phn: qun l cy nh phn ch cn qun l a ch nt gc:
Tree root;

struct TNode

{

DataType data;

TNode *pLeft, *pRight;

};

typedef TNode* Tree;

An array representation of a binary tree is not suitable for frequent insertions and deletions, even

though no storage is wasted if the binary tree is a complete binary tree. It makes insertion and deletion

in a tree costly. Therefore, instead of using an array representation, we can use a linked

representation, in which every node is represented as a structure with three fields

Chng 7: Cy (Tree)


*

a

b

c

d

e

g

h

i

f

j

k

Chng 7: Cy (Tree)

Binary Tree - Duyt cy nh phn

*
C 3 kiu duyt chnh c th p dng trn cy nh phn: Duyt theo th t trc - preorder (Node-Left-Right: NLR)Duyt theo th t gia - inorder (Left-Node-Right: LNR)Duyt theo th t sau - postorder (Left-Right-Node: LRN)Tn ca 3 kiu duyt ny c t da trn trnh t ca vic thm nt gc so vi vic thm 2 cy con
Chng 7: Cy (Tree)


Duyt theo th t trc NLR (Node-Left-Right)

Kiu duyt ny trc tin thm nt gc sau thm cc nt ca cy con tri ri n cy con phi

Th tc duyt c th trnh by n gin nh sau:

*

void NLR (Tree t)

{

if (t != NULL)

{

// X ly t tng ng theo nhu cu

NLR(t->pLeft);

NLR(t->pRight);

}

}

Chng 7: Cy (Tree)

Binary Tree - Duyt cy nh phn NLR

*

A

B

D

H

I

N

E

J

K

O

C

F

L

P

G

M

A

Kt qu:

B

D

H

I

N

E

J

O

K

C

F

L

P

G

M

Chng 7: Cy (Tree)


Duyt theo th t gia LNR (Left-Node-Right)
Kiu duyt ny trc tin thm cc nt ca cy con tri sau thm nt gc ri n cy con phiTh tc duyt c th trnh by n gin nh sau:
*

void LNR(Tree t)

{

if (t != NULL)

{

LNR(t->pLeft);

//X ly nut t theo nhu cu

LNR(t->pRight);

}

}

Chng 7: Cy (Tree)

Binary Tree - Duyt cy nh phn LNR

*

A

B

D

H

I

N

E

J

K

O

C

F

L

P

G

M

H

Kt qu:

D

N

I

B

J

O

E

K

A

F

P

L

C

M

G

Chng 7: Cy (Tree)


Duyt theo th t gia LRN (Left-Right-Node)
Kiu duyt ny trc tin thm cc nt ca cy con tri sau thm n cy con phi ri cui cng mi thm nt gcTh tc duyt c th trnh by n gin nh sau:
*

void LRN(Tree t)

{

if (t != NULL)

{

LRN(t->pLeft);

LRN(t->pRight);

// X ly tng ng t theo nhu cu

}

}

Chng 7: Cy (Tree)

Binary Tree - Duyt cy nh phn LRN

*

A

B

D

H

I

N

E

J

K

O

C

F

L

P

G

M

H

Kt qu:

N

I

D

O

J

K

E

B

P

L

F

M

G

C

A

Chng 7: Cy (Tree)

*
Mt v d quen thuc trong tin hc v ng dng ca duyt theo th t sau l vic xc nh tng kch thc ca mt th mc trn a

Chng 7: Cy (Tree)

*
Tnh ton gi tr ca biu thc da trn cy biu thc

(3 + 1)3/(9 5 + 2) (3(7 4) + 6) = 13

Chng 7: Cy (Tree)

Mt cch biu din cy nh phn khc

*
i khi, khi nh ngha cy nh phn, ngi ta quan tm n c quan h 2 chiu cha con ch khng ch mt chiu nh nh ngha phn trn. Lc , cu trc cy nh phn c th nh ngha li nh sau:
typedef struct tagTNode

{

DataTypeKey;

struct tagTNode*pParent;

struct tagTNode*pLeft;

struct tagTNode*pRight;

}TNODE;

typedef TNODE*TREE;

Chng 7: Cy (Tree)

Mt cch biu din cy nh phn khc

*

Chng 7: Cy (Tree)

Mt s thao tc trn cy
m s nodem s node lTnh chiu cao
*

Chng 7: Cy (Tree)

m s node

*

Chng 7: Cy (Tree)

m s node
S node (EmptyTree) = 0S node (Tree) = 1 + S node (Tree.Left)
+ S node (Tree.Right)

*

Chng 7: Cy (Tree)

m s node l

*

Chng 7: Cy (Tree)

m s node l

S nt l (EmptyTree) = 0

S nt l(RootOnly) = 1

S nt l(Tree) = S nt l (Tree.Left) +

S nt l (Tree.Right)

*

Chng 7: Cy (Tree)

Tnh chiu cao

*

Chng 7: Cy (Tree)

Tnh chiu cao

Height(Tree) = 1 + maximum(Height(Tree.Left), Height(Tree.Right))

Depth(EmptyTree) = 0

*

Chng 7: Cy (Tree)

Ni dung

*
Chng 7: Cy (Tree)

Binary Search Tree
Trong chng 6, chng ta lm quen vi mt s cu trc d liu ng. Cc cu trc ny c s mm do nhng li b hn ch trong vic tm kim thng tin trn chng (ch c th tm kim tun t)Nhu cu tm kim l rt quan trng. V l do ny, ngi ta a ra cu trc cy tha mn nhu cu trnTuy nhin, nu ch vi cu trc cy nh phn nh ngha trn, vic tm kim cn rt m hCn c thm mt s rng buc cu trc cy tr nn cht ch, d dng hnMt cu trc nh vy chnh l cy nh phn tm kim
Chng 7: Cy (Tree)

Binary Search Tree - nh ngha
Cy nh phn tm kim (CNPTK) l cy nh phn trong ti mi nt, kha ca nt ang xt ln hn kha ca tt c cc nt thuc cy con tri v nh hn kha ca tt c cc nt thuc cy con phiNh rng buc v kha trn CNPTK, vic tm kim tr nn c nh hngNu s nt trn cy l N th chi ph tm kim trung bnh ch khong log2NTrong thc t, khi xt n cy nh phn ch yu ngi ta xt CNPTK
*

Chng 7: Cy (Tree)

Binary Search Tree Vi du

*

44

18

88

13

37

59

108

15

23

40

55

71

Chng 7: Cy (Tree)

*


*

Chng 7: Cy (Tree)

*


*

Chng 7: Cy (Tree)

Cy nh phn v tm kim nh phn

*

0 1 2 3 4 5 6

34

41

56

63

72

89

95

Chng 7: Cy (Tree)


*

0 1 2 3 4 5 6

0 1 2

4 5 6

34

41

56

63

72

89

95

34

41

56

72

89

95

Chng 7: Cy (Tree)


*

0 1 2 3 4 5 6

0 1 2

4 5 6

0 2

4 6

34

41

56

63

72

89

95

34

41

56

72

89

95

34

56

72

95

Chng 7: Cy (Tree)

Binary Search Tree (BST)

*

63

41

89

34

56

72

95

Chng 7: Cy (Tree)

Binary Search Tree Biu din

*
Cu trc d liu ca CNPTK l cu trc d liu biu din cy nh phn ni chung (???)Thao tc duyt cy trn CNPTK hon ton ging nh trn cy nh phn (???)Ch : khi duyt theo th t gia, trnh t cc nt duyt qua s cho ta mt dy cc nt theo th t tng dn ca kha
Chng 7: Cy (Tree)

Binary Search Tree Duyt cy

*

Duyt inorder: 1 3 5 6 10 12 13 18 20 25 29 32 35 37 41 50

Duyt gia trn CNPTK

25

10

3

1

6

5

18

12

20

13

37

29

35

32

50

41

Chng 7: Cy (Tree)


*

Duyt postorder:

Duyt sau trn CNPTK

25

10

3

1

6

5

18

12

20

13

37

29

35

32

50

41

Chng 7: Cy (Tree)


*

Duyt preorder:

Duyt trc trn CNPTK

25

10

3

1

6

5

18

12

20

13

37

29

35

32

50

41

Chng 7: Cy (Tree)

Binary Search Tree Tm kim

*

25

10

3

1

6

5

18

12

20

13

37

29

35

32

50

41

Tm kim

13

Khc nhau

Ging nhau

Node gc nh hn

Node gc ln hn

Tm thy

S node duyt: 5

S ln so snh: 9

Tm kim trn CNPTK

Chng 7: Cy (Tree)


*

25

10

3

1

6

5

18

12

20

13

37

29

35

32

50

41

Tm kim

14

Khc nhau

Node gc nh hn

Node gc ln hn

Khng tm thy

S node duyt: 5

S ln so snh: 10

Tm kim trn CNPTK

Chng 7: Cy (Tree)

Tm mt phn t x trong CNPTK (dng quy):
*

TNode* searchNode(Tree T, DataType X)

{

if (T)

{

if(T->Key == X)

return T;

if(T->Key > X)

return searchNode(T->pLeft, X);

return searchNode(T->pRight, X);

}

return NULL;

}

Chng 7: Cy (Tree)

Tm mt phn t x trong CNPTK (dng vng lp):
*

TNode * searchNode(Tree T, DataType x)

{

TNode *p = T;

while (p != NULL)

{

if(x == p->Key) return p;

else

if(x < p->Key) p = p->pLeft;

elsep = p->pRight;

}

return NULL;

}

Chng 7: Cy (Tree)


*
Nhn xt:S ln so snh ti a phi thc hin tm phn t X l h, vi h l chiu cao ca cyNh vy thao tc tm kim trn CNPTK c n nt tn chi ph trung bnh khong O(log2n)
Chng 7: Cy (Tree)

Binary Search Tree Thm

*
Vic thm mt phn t X vo cy phi bo m iu kin rng buc ca CNPTKTa c th thm vo nhiu ch khc nhau trn cy, nhng nu thm vo mt nt ngoi s l tin li nht do ta c th thc hin qu trnh tng t thao tc tm kimKhi chm dt qu trnh tm kim cng chnh l lc tm c ch cn thmHm insert tr v gi tr:
1 khi khng b nh

0 khi gp nt c

1 khi thm thnh cng

Chng 7: Cy (Tree)

Thm mt phn t vo cy
*

int insertNode (Tree &T, DataType X)

{if (T) {

if(T->data == X) return 0;

if(T->data > X)

return insertNode(T->pLeft, X);

else

return insertNode(T->pRight, X);

}

T = new TNode;

if (T == NULL) return -1;

T->data = X;

T->pLeft = T->pRight = NULL;

return 1;

}

Chng 7: Cy (Tree)

*

6

V d to cy vi dy:
4, 6, 1, 2, 5, 7, 3

4

1

2

5

7

3

Chng 7: Cy (Tree)


*

30

12

49

51

17

22

56

70

68

65

Chng 7: Cy (Tree)

Binary Search Tree Hy mt phn t c kha X

*
Vic hy mt phn t X ra khi cy phi bo m iu kin rng buc ca CNPTKC 3 trng hp khi hy nt X c th xy ra:X l nt lX ch c 1 con (tri hoc phi)X c c 2 con
Chng 7: Cy (Tree)


*
Trng hp 1: X l nt l Ch n gin hy X v n khng mc ni n phn t no khc
44

18

88

13

37

59

108

15

23

40

55

71

T/h 1: huy X=40

Chng 7: Cy (Tree)

Trng hp 2: X ch c 1 con (tri hoc phi) Trc khi hy X ta mc ni cha ca X vi con duy nht ca n
*

44

18

88

13

37

59

108

15

23

55

71

T/h 2: huy X=37

Trc khi huy X ta moc noi cha cua X vi con duy nhat cua no

Chng 7: Cy (Tree)

Trng hp 3: X c 2 con:Khng th hy trc tip do X c 2 con Hy gin tip: Thay v hy X, ta s tm mt phn t th mng Y. Phn t ny c ti a mt conThng tin lu ti Y s c chuyn ln lu ti XSau , nt b hy tht s s l Y ging nh 2 trng hp uVn : chn Y sao cho khi lu Y vo v tr ca X, cy vn l CNPTK
*

Chng 7: Cy (Tree)

Trng hp 3: X c 2 con:C 2 phn t tha mn yu cu: Phn t tri nht trn cy con phiPhn t phi nht trn cy con triVic chn la phn t no l phn t th mng hon ton ph thuc vo thch ca ngi lp trnh y, ta s chn phn t phi nht trn cy con tri lm phn t th mng
*

Chng 7: Cy (Tree)

Trng hp 3: X c 2 con:Khi hy phn t X=18 ra khi cy, phn t 23 l phn t th mng:
*

44

18

88

13

37

59

108

15

23

40

55

71

T/h 3: huy X=18

30

23

Chng 7: Cy (Tree)

Trng hp 3: X c 2 con:Hm delNode tr v gi tr 1, 0 khi hy thnh cng hoc khng c X trong cy:
int delNode(Tree &T, DataType X)
Hm searchStandFor tm phn t th mng cho nt p
void searchStandFor(Tree &p, Tree &q)

*

Chng 7: Cy (Tree)

*


int delNode(Tree &T, DataType X)

{

if (T == NULL)return 0;

if (T->data > X)return delNode(T->pLeft, X);

if (T->data < X)return delNode(T->pRight, X);

TNode* p = T;

if (T->pLeft == NULL)

T = T->pRight;

else

if (T->pRight == NULL) T = T->pLeft;

else // T c 2 con

searchStandFor(p, T->pRight);

delete p;

}

Chng 7: Cy (Tree)

Tm phn t th mng
*

void searchStandFor(Tree &p, Tree &q)

{

if (q->pLeft)

searchStandFor(p, q->pLeft);

else

{

p->data = q->data;

p = q;

q = q->pRight;

}

}

Chng 7: Cy (Tree)


*
V d xa 51:
Chng 7: Cy (Tree)


*
V d xa 83:
Chng 7: Cy (Tree)


*
V d xa 36:
Chng 7: Cy (Tree)


*
Xa nt gc (2 ln):
Chng 7: Cy (Tree)


*
V d xa 15:
Chng 7: Cy (Tree)


*
42 l th mng
Chng 7: Cy (Tree)


*

Chng 7: Cy (Tree)


*
Kt qu xo ln 1:
Chng 7: Cy (Tree)


*
V d xa 42
Chng 7: Cy (Tree)


*

Chng 7: Cy (Tree)


*

Chng 7: Cy (Tree)


*
Xa 15, sau 42:
Chng 7: Cy (Tree)

Binary Search Tree Hy ton b cy
Vic ton b cy c th c thc hin thng qua thao tc duyt cy theo th t sau. Ngha l ta s hy cy con tri, cy con phi ri mi hy nt gc
*

void removeTree(Tree &T)

{

if(T)

{

removeTree(T->pLeft);

removeTree(T->pRight);

delete(T);

}

}

Chng 7: Cy (Tree)

Binary Search Tree

*
Nhn xt:Tt c cc thao tc searchNode, insertNode, delNode u c phc tp trung bnh O(h), vi h l chiu cao ca cy Trong trong trng hp tt nht, CNPTK c n nt s c cao h = log2(n). Chi ph tm kim khi s tng ng tm kim nh phn trn mng c th tTrong trng hp xu nht, cy c th b suy bin thnh 1 danh sch lin kt (khi m mi nt u ch c 1 con tr nt l). Lc cc thao tc trn s c phc tp O(n)V vy cn c ci tin cu trc ca CNPTK t c chi ph cho cc thao tc l log2(n)
Chng 7: Cy (Tree)

Binary Search Tree

*
1,2,3,4,5
1

2

3

4

5

Chng 7: Cy (Tree)

Ni dung

*
Chng 7: Cy (Tree)

AVL Tree - nh ngha

*
Cy nh phn tm kim cn bng l cy m ti mi nt ca n cao ca cy con tri v ca cy con phi chnh lch khng qu mt.
Chng 7: Cy (Tree)

AVL Tree V d

*

44

23

88

13

37

59

108

15

30

40

55

71

Chng 7: Cy (Tree)

AVL Tree

*
Lch s cy cn bng (AVL Tree): AVL l tn vit tt ca cc tc gi ngi Nga a ra nh ngha ca cy cn bng Adelson-Velskii v Landis (1962)T cy AVL, ngi ta pht trin thm nhiu loi CTDL hu dng khc nh cy -en (Red-Black Tree), B-Tree, Cy AVL c chiu cao O(log2(n))
Chng 7: Cy (Tree)

AVL Tree
Ch s cn bng ca mt nt: nh ngha: Ch s cn bng ca mt nt l hiu ca chiu cao cy con phi v cy con tri ca n. i vi mt cy cn bng, ch s cn bng (CSCB) ca mi nt ch c th mang mt trong ba gi tr sau y: CSCB(p) = 0 cao cy tri (p) = cao cy phi (p)CSCB(p) = 1 cao cy tri (p) < cao cy phi (p)CSCB(p) =-1 cao cy tri (p) > cao cy phi (p) tin trong trnh by, chng ta s k hiu nh sau: p->balFactor = CSCB(p); cao cy tri (p) k hiu l hL cao cy phi(p) k hiu l hR
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Chng 7: Cy (Tree)

AVL Tree Biu din

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#defineLH-1/* Cay con trai cao hn */

#defineEH 0/* Hai cay con bang nhau */

#defineRH 1/* Cay con phai cao hn */

struct AVLNode{

charbalFactor; // Ch s cn bng

DataTypedata;

tagAVLNode*pLeft;

tagAVLNode*pRight;

};

typedef AVLNode*AVLTree;

Chng 7: Cy (Tree)

AVL Tree Biu din

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Trng hp thm hay hy mt phn t trn cy c th lm cy tng hay gim chiu cao, khi phi cn bng li cyVic cn bng li mt cy s phi thc hin sao cho ch nh hng ti thiu n cy nhm gim thiu chi ph cn bngCc thao tc c trng ca cy AVL:Thm mt phn t vo cy AVLHy mt phn t trn cy AVLCn bng li mt cy va b mt cn bng
Chng 7: Cy (Tree)

AVL Tree
Cc trng hp mt cn bng:Ta s khng kho st tnh cn bng ca 1 cy nh phn bt k m ch quan tm n cc kh nng mt cn bng xy ra khi thm hoc hy mt nt trn cy AVL Nh vy, khi mt cn bng, lch chiu cao gia 2 cy con s l 2C 6 kh nng sau: Trng hp 1 - Cy T lch v bn tri : 3 kh nngTrng hp 2 - Cy T lch v bn phi: 3 kh nng
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Chng 7: Cy (Tree)

AVL Tree
Trng hp 1: cy T lch v bn tri
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T

T1

L1

R1

h

h-1

h-1

L

R1

T

T1

L1

h

h-1

L

R

R

h-1

T

T1

L1

h

h-1

h

L

R

L1

Chng 7: Cy (Tree)

AVL Tree
Trng hp 2: cy T lch v bn phi
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T

h-1

R1

T1

h

h

R

L

L1

T

T1

L1

R1

h

h-1

R

T

R1

T1

L1

h

h-1

R

L

h-1

L

h-1

Chng 7: Cy (Tree)

AVL Tree

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Cc trng hp mt cn bng:Cc trng hp lch v bn phi hon ton i xng vi cc trng hp lch v bn tri. V vy, ch cn kho st trng hp lch v bn tri. Trong 3 trng hp lch v bn tri, trng hp T1 lch phi l phc tp nht. Cc trng hp cn li gii quyt rt n gin.
Chng 7: Cy (Tree)

AVL Tree - Cn bng li cy AVL

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T/h 1.1: cy T1 lch v bn tri. Ta thc hin php quay n Left-Left
T

T1

L1

R1

h

h-1

h-1

L

R

T1

T

L1

R1

h

h-1

R

h-1

Chng 7: Cy (Tree)


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T/h 1.2: cy T1 khng lch. Ta thc hin php quay n Left-Left
T

T1

L1

h

h-1

h

L

R

T1

T

L1

h

R

h-1

R1

R1

h

Chng 7: Cy (Tree)


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T/h 1.3: cy T1 lch v bn phi. Ta thc hin php quay kp Left-Right Do T1 lch v bn phi ta khng th p dng php quay n p dng trong 2 trng hp trn v khi cy T s chuyn t trng thi mt cn bng do lch tri thnh mt cn bng do lch phi ? cn p dng cch khc
Chng 7: Cy (Tree)

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R1

T

T1

L1

h

h-1

L

R

h-1

T

T1

L1

h-1

R1

R

h-1

T2

L2

R2

T

T1

L1

h-1

R

h-1

T2

L2

R2

Chng 7: Cy (Tree)

Lu : Trc khi cn bng cy T c chiu cao h+2 trong c 3 trng hp 1.1, 1.2 v 1.3Sau khi cn bng:Trng hp 1.1 v 1.3 cy c chiu cao h+1Trng hp 1.2 cy vn c chiu cao h+2. y l trng hp duy nht sau khi cn bng nt T c c ch s cn bng 0Thao tc cn bng li trong tt c cc trng hp u c phc tp O(1)
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Chng 7: Cy (Tree)


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T/h 1.1: cy T1 lch v bn tri. Ta thc hin php quay n Left-Left
T

T1

L1

R1

h

h-1

h-1

L

R

T1

T

L1

R1

h

h-1

R

h-1

Chng 7: Cy (Tree)


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void rotateLL(AVLTree &T) //quay n Left-Left

{

AVLNode* T1 = T->pLeft;

T->pLeft= T1->pRight;

T1->pRight= T;

switch(T1->balFactor) {

case LH: T->balFactor = EH;

T1->balFactor = EH;

break;

case EH: T->balFactor = LH;

T1->balFactor = RH;

break;

}

T= T1;

}
Quay n Left-Left:
Chng 7: Cy (Tree)


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Quay n Right-Right:
void rotateRR (AVLTree &T)//quay n Right-Right

{

AVLNode*T1 = T->pRight;

T->pRight= T1->pLeft;

T1->pLeft= T;

switch(T1->balFactor){

case RH: T->balFactor = EH;

T1->balFactor= EH;

break;

case EH: T->balFactor = RH; T1->balFactor= LH;

break;

}

T = T1;

}

Chng 7: Cy (Tree)


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Quay kp Left-Right:
void rotateLR(AVLTree &T)//quay kp Left-Right

{AVLNode*T1 = T->pLeft;

AVLNode*T2 = T1->pRight;

T->pLeft= T2->pRight;

T2->pRight= T;

T1->pRight= T2->pLeft;

T2->pLeft= T1;


case LH: T->balFactor = RH; T1->balFactor = EH; break;

case EH: T->balFactor = EH; T1->balFactor = EH; break;

case RH: T->balFactor = EH; T1->balFactor = LH; break;

}

T2->balFactor = EH;

T = T2;

}

Chng 7: Cy (Tree)


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Quay kep Right-Left
void rotateRL(AVLTree &T)//quay kp Right-Left

{AVLNode*T1 = T->pRight;

AVLNode*T2 = T1->pLeft;

T->pRight= T2->pLeft;

T2->pLeft= T;

T1->pLeft= T2->pRight;

T2->pRight= T1;


case RH: T->balFactor = LH; T1->balFactor = EH; break;

case EH: T->balFactor = EH; T1->balFactor = EH; break;

case LH: T->balFactor = EH; T1->balFactor = RH; break;

}

T2->balFactor = EH;

T = T2;

}

Chng 7: Cy (Tree)


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Cn bng khi cy b lch v bn tri:
int balanceLeft(AVLTree &T)

//Cn bng khi cy b lch v bn tri

{

AVLNode*T1 = T->pLeft;


case LH:rotateLL(T); return 2;

case EH:rotateLL(T); return 1;

case RH:rotateLR(T); return 2;

}

return 0;

}

Chng 7: Cy (Tree)


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Cn bng khi cy b lch v bn phi
int balanceRight(AVLTree &T )

//Cn bng khi cy b lch v bn phi

{

AVLNode*T1 = T->pRight;


case LH:rotateRL(T); return 2;

case EH:rotateRR(T); return 1;

case RH:rotateRR(T); return 2;

}

return 0;

}

Chng 7: Cy (Tree)

AVL Tree - Thm mt phn t trn cy AVL
Vic thm mt phn t vo cy AVL din ra tng t nh trn CNPTKSau khi thm xong, nu chiu cao ca cy thay i, t v tr thm vo, ta phi ln ngc ln gc kim tra xem c nt no b mt cn bng khng. Nu c, ta phi cn bng li nt nyVic cn bng li ch cn thc hin 1 ln ti ni mt cn bng Hm insertNode tr v gi tr 1, 0, 1 khi khng b nh, gp nt c hay thnh cng. Nu sau khi thm, chiu cao cy b tng, gi tr 2 s c tr v
int insertNode(AVLTree &T, DataType X)

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Chng 7: Cy (Tree)



{intres;

if (T)

{if (T->key == X) return 0; // c

if (T->key > X)

{ res= insertNode(T->pLeft, X);

if(res < 2) return res;

switch(T->balFactor)

{case RH: T->balFactor = EH; return 1;

case EH: T->balFactor = LH; return 2;

case LH: balanceLeft(T); return 1;

}

}

......................................................

}

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insertNode2

Chng 7: Cy (Tree)



{

......................................................

else // T->key < X

{ res= insertNode(T-> pRight, X);



{case LH: T->balFactor= EH; return 1;

case EH: T->balFactor= RH; return 2;

case RH: balanceRight(T); return 1;

}

}

......................................................

}

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insertNode3

Chng 7: Cy (Tree)



{

......................................................

T= new TNode;

if(T == NULL) return -1; //thiu b nh

T->key = X;

T->balFactor= EH;

T->pLeft = T->pRight = NULL;

return 2; // thnh cng, chiu cao tng

}

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Chng 7: Cy (Tree)

AVL Tree - Hy mt phn t trn cy AVL
Cng ging nh thao tc thm mt nt, vic hy mt phn t X ra khi cy AVL thc hin ging nh trn CNPTKSau khi hy, nu tnh cn bng ca cy b vi phm ta s thc hin vic cn bng liTuy nhin vic cn bng li trong thao tc hy s phc tp hn nhiu do c th xy ra phn ng dy chuynHm delNode tr v gi tr 1, 0 khi hy thnh cng hoc khng c X trong cy. Nu sau khi hy, chiu cao cy b gim, gi tr 2 s c tr v:
int delNode(AVLTree &T, DataType X)

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Chng 7: Cy (Tree)



{intres;

if(T==NULL)return 0;

if(T->key > X)

{ res = delNode (T->pLeft, X);

if(res < 2)return res;


{ case LH: T->balFactor = EH; return 2;

case EH: T->balFactor = RH; return 1;

case RH: return balanceRight(T);

}

} // if(T->key > X)

......................................................

}

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delNode2

Chng 7: Cy (Tree)



{

......................................................

if(T->key < X)

{ res= delNode (T->pRight, X);



{ case RH: T->balFactor = EH; return 2;


case LH: return balanceLeft(T);

}

} // if(T->key < X)

......................................................

}

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delNode3

Chng 7: Cy (Tree)



{......................................................

else//T->key == X

{ AVLNode*p = T;

if(T->pLeft == NULL) { T = T->pRight; res = 2; }

else if(T->pRight == NULL) { T = T->pLeft; res = 2; }

else //T c c 2 con

{ res = searchStandFor(p,T->pRight);



{ case RH: T->balFactor = EH; return 2;


case LH: return balanceLeft(T);

}

}

delete p; return res;

}

}

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Chng 7: Cy (Tree)


int searchStandFor(AVLTree &p, AVLTree &q)

//Tm phn t th mng

{intres;

if(q->pLeft)

{ res= searchStandFor(p, q->pLeft);


switch(q->balFactor)

{ case LH: q->balFactor = EH; return 2;

case EH: q->balFactor = RH; return 1;

case RH: return balanceRight(T);

}

} else

{ p->key = q->key; p = q; q = q->pRight; return 2;

}

}

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Chng 7: Cy (Tree)

AVL Tree

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Nhn xt:Thao tc thm mt nt c phc tp O(1)Thao tc hy mt nt c phc tp O(h)Vi cy cn bng trung bnh 2 ln thm vo cy th cn mt ln cn bng li; 5 ln hy th cn mt ln cn bng li
Chng 7: Cy (Tree)

AVL Tree
Nhn xt:Vic hy 1 nt c th phi cn bng dy chuyn cc nt t gc cho n phn t b hy trong khi thm vo ch cn 1 ln cn bng cc b di ng tm kim trung bnh trong cy cn bng gn bng cy cn bng hon ton log2n, nhng vic cn bng li n gin hn nhiuMt cy cn bng khng bao gi cao hn 45% cy cn bng hon ton tng ng d s nt trn cy l bao nhiu
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