BinarySearchTr ADTtedtoSes : unordered collection of values Iobjects - - PowerPoint PPT Presentation

SLIDE 1

BinarySearchTr

SLIDE 2

ADTtedtoSes

• Set

: unordered collection of valuesIobjects insert H) ; memberlx) ; remove 4) ; emptyD; sized, dearD; (find LX ))

• We call the values we store keys

,

• We assume the keys are from some
• rdered sets

ie , for any two keys ix.yes , we have exactly

• ne of

day , x=y , ya x

• Want implementations where all operations

are as efficientHast as possible

• Compare with
• ur list implementations

where we could have some operations fast , but others slow Q : What will count as " fast " ?

SLIDE 3

someRelatedlontainerADTSMottiset.li

he set, but with multiplicities (aka beg) Count(x) Map : unordered collection of Lkey, take pairs , associating at most one valve with each key . put Ikey,ral ) 4in place of insert × get Ckey) Arcturus value associated with key

Dictionary

: like map , but associates a collection

• f

values with each key .

• Implementations of these

are simple extortions to implementations of sets , which we focus

• n .

SLIDE 4

searchtrees.czoat.SI

, Multi set , Mgp , . . . ADTS with (at least) insert,

find , removes

, all

efficient

• ff¥¥im
• by

where n= # keys in the set like binary search

( Our first attempt will not

succeed at

• giving 0110g n) operations

.)

SLIDE 5

BinarySearohT

A

BST

is

• a binary tree
• with

nodes labelled by keys

• for every two

nodes v.v : ' if u is in the left subtree of v then key (a) a key 4)

it

III.rig

btreeoe} BST

SLIDE 6

Ex

= ⑤

• ⑤

←T

• %

⑤r

'

③

⇐

SLIDE 7

i

¥6

④

SLIDE 8 W

Ey : Reasoning with the

µ

• rder invariant .

/

↳ v3

1.1

.

\v2

vi.T

• LI

key Cut) a key Cu)

key (re) > key Cu) key Crs)

? key Cu)

Kayla)

a key ( w ) a key Cvs) ⇒ key Cnn) a key Los)

SLIDE 9

BSTmemberkdpseudo-codef-indlt.ir

) 4 return true if t appears

in { 4 subtree rooted at it . if t a key G) ! v has a left subtree return find Lt , left Iv)) if t > key Cr ) 4 v has a right subtree return find (t, right Cr )) if key 4) = t return true return false } member(t) return true ift t is in the tree . } return findIt , root)

SLIDE 10

BSTfindltivlpseudo-code-alternater-er.in

find Lt , r) 4 return true if t appears in { 4 subtree rooted at it . if key 4) = t return true if t a key G) ! v has a left subtree return find Lt , left Iv)) if t > key ( r ) I v has a right subtree return find (t , right Cr )) return false }

• Oi. Which version is better ?

Since keyG) =t will almost always be false , the first version should do fewer comparisons , so be faster .

SLIDE 11

BSTinserttxlpseudo.co#

insert Had ds t to the tree Hassomes t is not in the tree . already a ← node at which find It,root) terminates if t a key (a) give u a new left child with key t else give m a new right child with key t . }

SLIDE 12

BSTInsertExamp

insertthy insert

2

←③

7

%

SLIDE 13

BST

insert4) Pseudo - code

• alternate version

f

• insert Lt , DX insert t in the subtree rooted at v , if it is not there

. { if t a key G) ! v has a left subtree insert Lt , left Iv)) if t > key Cr ) I v has a right subtree insert Ct, right GD if t a key G) 11 here v has no left child give v a new left child with key t if key 4) at it t > key 4) 4 here r has no right child return true give r a new right child with key t . } Kit we reach here , t= keylvl , so do nothing . insert It){ Dodds t to the tree ,if it is not already there. insert It , root )

}

SLIDE 14

Ini¥BTiExamp

1) e. start with an empty . in tae given order insert

E ? ' Ii '

←

2) ' start with an empty BST

• insert

1,213,516,78 ' in the

• rder given
• ②-③-⑤-⑥-⑨
• ⑧

SLIDE 15

BST

remove It)

• We

consider 3 cases , of increasing difficulty .

• Catisatal

i) find the node r wi ii) delete ✓ 'th key G) = t

E

remove4),¥⑤

⑤

F

② * "

⑨

L

SLIDE 16

BSTremoretttfki.tisatanodewithtchild.is

find the node V with key Cut = t let u be the child of r iii) replace v with the subtree rooted at a Exampte : remove(3) , then remove4)

→ ←

¥7

1-③

¥,

⑤

⑧

SLIDE 17

E×amp BST removeIt)

where nodeft) has 1 child remove

③

T

• o ⇒ of '⑧

'

⑥

• eCT

*

SLIDE 18

BST remove : Case 3 Preparation

• In an ordered set

f =L . . . Si- I , Si , Siti , sit 2 . . . 7 Sin is the predecessor of si Sit , is the successor

• f

Si

• Let s

be the set of keys in the Best , and t be

• ur target

K .

• We will want to locate the predecessor/

successor of Tins.

SLIDE 19

BST remove : Case 3 Preparation

• 0bse An

in- order traversal of a BST visits nodes in the order of their keys .

• Let 4=4 , vz .

. . . un) be the set of nodes in the BST in the order visited by an in - order traversal

• For an
• rdered set

X , and xtX , write succxtx) for the successor of × in X.

• Then we have

a) if key (a) =k then key ( succubi)) = succslk) b) If nodeck) =u then node(sudsCKD = such

SLIDE 20

BSTremoveicasezpreparation.IT

node v has a right child , it is easy to find its successor : 4.↳ ← succor) is the first

\

node visited by an

• in - order traversal
• f the right subtree
• f v.

E%f

→ such

§

µ

Succor)

4.

v ⇒

F.

I

Do

T

• such#↳

to skew)

SLIDE 21

smallest key

in a B5T .

→

¥884

¥0

SLIDE 22

BSTremoveilase3preparation.TO

find the successor of node v (assuming v has a right child) use: suck) { u⇐ right G) while ( left Cm ) exists) { } U * left Cns return a } The next node after r contains the next key after keyG)

• i.e. the

smallest key that is larger than keyG) .

SLIDE 23

Bs Trema#

Casey

: t is at a node with 2 children i) find the node v with key G) =t ii ) find the successor of V

• call it

u . iii) key f) ← key G) iv) delete re : a) if u is a leaf , delete it b) if u is not a leaf , it has

• ne child

w replace u with the subtree rooted at w

(as

in Case 2)

SLIDE 24

BSTremovelhwhennodelldhas2chddreurem.ve

(5)7¥

• µ⑤

⑤

,

€ T

• ← ¥

µ? '

①

¥10

SLIDE 25

BSeCk)whenhdren.

removed

,②T⑧D

"

A

④#

"I

⇒ A

,④,

A

A

• 40 ✓

replace

⑤ ④

with ⑨

A

SLIDE 26

Complexity

• f B5T

Operations # keys

• Measure

as a function of : height 1h)

• r site Ln) :
• All operations essentially

involve traversing a path from the root to a node v , where v in the worst case it is a leaf of maximum depth

• so :

member :O (h) , 0 (n)

/ T

• ÷:::÷:÷:

remove :O ( h) , old 'h= ? eg.tt) his small relative to n .

• For

"tall skinny " trees leg Tbh is proportional to n .

⇐ %.IS?I3sBshIEe

' r ' 90

SLIDE 27

• A perfeutbinarytree of height

h is a binary tree

• f height h with the

Max . number of nodes :

• I

1ha

T

• .

✓ I

r

an £5

✓

DAH ha

X

I

↳

→

×

SLIDE 28

SLIDE 29

Notice :

Because a perfect binary tree of height h is like this: 2h -I

÷:/

2h leaves 2" t 2h - I = 2. 2h

• I

= 2 htt . I

SLIDE 30 Claim : For every set S of n keys , there exists a BST for S with height at most It Logan Pro¥ Let h be the smallest integer st . 2h > n , and m=2h . So : 2h In > 2h

• I

log, 2" slogan > log,2h" h =

logznyh-lha.lt

Logan

• let T be the perfect binary tree of

height h= login . label the first n nodes of T ( as visited by an in - order traversal) with the keys of 5 , and delete the remaining nodes (to get T ' ) .

• T

' is a B5T for S with height h = log. m < It login . So , there is always a B5T with height

• log n

SLIDE 31

0ptimalBSTTnsertionorder-G.ir

en a set

• f

keys , we can insert them so as to get a minimum height BST : Consider : 0 ← What can we say about

• / \

the key here?

( It

is the

← to

6 to

median ← to 6 to 610

68

key ) Observe : The first key inserted into a BST is at the root forever

( unless

we remove it from the BST) ⇒

SLIDE 32

Optimal BST Insertion Order .

¥→

I

D

* AI1II

seethe

median key " principle

SLIDE 33

• So , there

is always a B5T with height

• log n
• Can

we maintain min . height with Ollogn) insert 4 remove ? . Consider : insert I ⑤

• ← TO

⇒ ② ⑥

←

6

66

⑤ T

• ⑦

④

' B is the only min height B5T for 1 . . 7.

• A → B required

" moving every node "

• To get 0( log

n)

• perations , we need another

kind of search tree , other than plain BSTS .

SLIDE 34

To get efficient

search trees , give up at least

• ne of :
• binary
• min height

.to/logn)operatims-

SLIDE 35

Ends

SLIDE 36

Testt

• Coverage

: lectures up to last friday t At. . Short answer

• r exercises

Focus : data structures - properties %

• perations
• Note:

use versions from lectures . F⇒ . BST remove cqn use pied OR a .

D.

D

pied Sufc

• Usetheouefromclas-webp.ge
• Slides coming