Lecture12 Updated

8/9/2019 Lecture12 Updated

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Lecture #12

•

Binary Tree Traversals• Using Binary Trees to Evaluate Expressions

• Binary Search Trees• Binary Search Tree Operations

– Searching for an item – Inserting a ne item – !in"ing the minimum an" maximum items – rinting out the items in or"er –

$eleting the hole tree


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Binary Tree Traversals%hen e process all the no"es in a tree& it's calle" a

traversal(

There are four common ays to traverse a tree(

1( re)or"er traversal2( In)or"er traversal*( ost)or"er traversal

+( Level)or"er traversal

Let's see a pre)or"er traversal first,


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The reor"er Traversal

reor"er- 1( rocess the current no"e( 2( rocess the no"es in the

left su.)tree( *( rocess the no"es in the

right su.)tree(

By /process the current no"e0 e typically mean one ofthe folloing-

1( rint the current no"e's value out(2( Search the current no"e to see if its value matches

the one you're searching for(*( "" the current no"e's value to a total for the tree+( Etc

NULL

/a0

/.0 /c0NULL

/"0NULL NULL NULL /e0NULL

root


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voi" reOr"er34o"e 5cur67

if 3cur 88 4ULL6 99 if empty& return return:

cout ;; cur)


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cout ;; cur)


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cout ;; cur)


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cout ;; cur)


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cout ;; cur)


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cout ;; cur)


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The In)or"er Traversal1( rocess the no"es in the left

su.)tree(2( rocess the current no"e(*( rocess the no"es in the right

su.)tree(

voi" InOr"er34o"e 5cur67


InOr"er3cur)


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The In)or"er Traversal1( rocess the no"es in the left

su.)tree(2( rocess the current no"e(*( rocess the no"es in the right

su.)tree(



InOr"er3cur)


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The ost)or"er Traversal1( rocess the no"es in the left

su.)tree(2( rocess the no"es in the right

su.)tree(*( rocess the current no"e(

voi" ostOr"er34o"e 5cur67


ostOr"er3cur)


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The ost)or"er Traversal1( rocess the no"es in the left

su.)tree(2( rocess the no"es in the right

su.)tree(*( rocess the current no"e(

voi" ostOr"er34o"e 5cur67


ostOr"er3cur)


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The Level Or"er TraversalIn a level or"er traversal e visit each level's no"es& from

left to right& .efore visiting no"es in the next level(

NULL

/a0

/.0 /c0

/"0NULL NULL NULL

/e0NULL

root

>ere's the algorithm-

1( Use a temp pointer varia.le an"

a ?ueue of no"e pointers(2( Insert the root no"e pointer

into the ?ueue(*( %hile the ?ueue is not empty-

( $e?ueue the top no"epointer an" put it in temp(B( rocess the no"e(@( "" the no"e's chil"ren to

?ueue if they are not 4ULL(

front rear

A

A2 AC

C AD

720780

800760

A

A

temp A

a

A2AC

A2

AC

.

CAD

AC

CAD

c

NULL

/f0NULL

900

C

AD

" Etc


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Traversal Overvie& art 1

NULL

/a0

/.0 /c0

NULL

/"0NULL NULL NULL

/e0NULL

root

re)or"er

#1

#2

#*

#+

#F

#C ##1

#1*

#1+

#1F

#1D

#1

NULL

/a0

/.0 /c0

NULL

/"0

NULL NULL NULL

/e0

NULL

root

In)or"er

#1+

#1

#A

#2

#+

#D #C#1

#12

#1* #1F #1A

#1

#D #A #11 #12

#1A #1C

#* #F # #11

#1D #1C

1( rocess current no"e

2( Traverse left*( Traverse right

1( Traverse left

2( rocess current no"e*( Traverse right


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Traversal Overvie& art 2

NULL

/a0

/.0 /c0

NULL

/"0NULL NULL NULL

/e0NULL

root

ost)or"er

#1

#1

#12

#2

#F

#D #A#1

#11

#1*

#1+

#1A

#1C

NULL

/a0

/.0 /c0

NULL

/"0

NULL NULL NULL

/e0

NULL

root

Level)or"er

#1

#2

#+#F

#*

#* #+ #C #

#1F #1D

1( Traverse left2( Traverse right*( rocess current no"e


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Big)Oh of TraversalsG

Huestion- %hat're the .ig)ohs of each of our traversalsG

nser- %ell& since a traversal must visit each no"eexactly once

an" since there are n no"es in a tree

the .ig)oh for any of the traversals is

O3n6


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Traversal @hallenge

@hallenge- %hat or"er ill thefolloing no"es .e printe" out ife use an in)or"er traversalG

NULL

/Larry0

/!ran0 /on"a0

/$anny0NULL NULL NULL

/JacK0NULL

root

/Tom0NULL

/Sam0NULL NULL

• The class ill split intoleft an" right teams

• One stu"ent from eachteam ill come up to the

.oar"• Each stu"ent can either

– rite one ne item or – fix a single error in

their teammatessolution

• Then the next to peoplecome up& etc(

• The team that completes

their program first ins,

ULES


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Expression Evaluation%e can represent arithmetic expressions using a

.inary tree(

ptr

!or example& the tree onthe left represents the

expression- 3FD653*)16

Once you have an expressionin a tree& its easy to

evaluate it an" get theresult(

Let's see ho,


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Expression Evaluation

1( If the current no"e is a

num.er& return its value(

2( ecursively evaluate the leftsu.tree an" get the result(

*( ecursively evaluate the right

su.tree an" get the result(+( pply the operator in the

current no"e to the left an"right results: return the

result(

ptr

3FD653*)16

>ere's our evaluation function( %e start .y passing in a

pointer to the root of the tree(

cur


num.er& return its value(2( ecursively evaluate the left

su.tree an" get the result(

*( ecursively evaluate the rightsu.tree an" get the result(

+( pply the operator in thecurrent no"e to the left an"right results: return theresult(

cur

1( If the current no"e is anum.er& return its value(




c u r


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result(

ptr

3FD653*)16



cur






cur

esult 8 F

1( If the current no"e is anum.er& return its value(




c u r


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result(

ptr

3FD653*)16



cur






cur

esult 8 F

esult 8 D

FD 8 11


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result(

ptr

3FD653*)16



cur

*)1 8 2

esult 8 11






c u result 8 *

esult 8 1


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result(

ptr

3FD653*)16



cur

*)1 8 2

esult 8 11

esult 8 2

1152822

The result is 22(


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result(



Huestion- %hich otheralgorithm "oes this remin"

you ofG


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Binary Search Trees

Binary Search Trees are a type of .inary tree ithspecific properties that maKe them very efficient to

search for a value in the tree(

LiKe regular Binary Trees&e store an" search forvalues in Binary Search

Trees

>ere's an example BST

NULL

/Larry0

/!ran0 /on"a0


/JacK0NULL

root

/Tom0NULL

/Sam0NULL NULL


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Binary Search TreesHuestion- %hich of the folloing are vali" BSTsG

NULL

/Larry0

/!ran0 /on"a0


/4icK0NULL

NULL

/Larry0

/!ran0

/$anny0NULL

NULL

/lex0NULL NULL

NULL

/Nanny0

/my0 /4icK0

NULL

/Na""y0NULL

NULLNULLNULL


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Operations on a Binary Search Tree

• $etermine if the .inary search tree is empty• Search the .inary search tree for a value• Insert an item in the .inary search tree• $elete an item from the .inary search tree• !in" the height of the .inary search tree• !in" the num.er of no"es an" leaves in the

.inary search tree• Traverse the .inary search tree• !ree the memory use" .y the .inary search tree

>ere's hat e can "o to a BST-

hi B T


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Searching a BSTInput- value to search for

Output- TUE if foun"& !LSE otheriseStart at the root of the treePeep going until e hit the 4ULL pointer

If is e?ual to current no"e's value& then foun", If is less than current no"e's value& go left If is greater than current no"e's value& go right

If e hit a 4ULL pointer& not foun"(

NULL

/Larry0

/!ran0 /on"a0

NULL NULL NULL /Mary0NULL

NULL

/Barry0

Let's search for

Mary(

Mary 88 LarryGGMary ; LarryGGMary 88 !ranGGMary ; !ranGGMary < !ranGGMary 88 MaryGG

S hi BST


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Searching a BSTStart at the root of the tree

Peep going until e hit the 4ULL pointer If is e?ual to current no"e's value& then foun", If is less than current no"e's value& go left If is greater than current no"e's value& go right

If e hit a 4ULL pointer& not foun"(

Sho ho to search for-

1( Phang2( $ale*( Sam


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Searching a BST >ere are to "ifferent BST search algorithms in @&

one recursive an" one iterative-

bool Search(int V, Node *ptr)

{

if (ptr == NULL)return(false); // nope

else if (V == ptr-!alue)

return(true); // found"""

else if (V # ptr-!alue)

return(Search(V,ptr-left));

elsereturn(Search(V,ptr-ri$ht));

%

bool Search(int V,Node *ptr)

{

&hile (ptr "= NULL)

{ if (V == ptr-!alue)

return(true);


ptr = ptr-left;

else

ptr = ptr-ri$ht; %

return(false); // nope

%

Let's trace through the recursive version


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ecursive BST SearchLets search for 1+(


{

if (ptr == NULL)


else if (V == ptr-!alue) return(true); // found"""


return(Search(V, ptr-left));

else

return(Search(V, ptr-ri$ht));

%

1*

A 1A

NULL NULL

NULLNULL

*NULLNULL

1+ 1

NULL

poot

!oid 'ain(!oid)

{

bool bnd;

bnd = Search(,p+oot);

%

ptr)


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{

if (ptr == NULL)





else


%

1*

A 1A

NULL NULL

NULLNULL

*NULLNULL

1+ 1

NULL

poot

!oid 'ain(!oid)

{

bool bnd;


%

ptr)<

bool Search(int V, Node *ptr){

if (ptr == NULL)


else if (V == ptr-!alue)

return(true); // found"""

else if (V # ptr-!alue) return(Search(V, ptr-left));

else


%

ptr)<

true

true

true


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{

if (ptr == NULL)





else


%

1*

A 1A

NULL NULL

NULLNULL

*NULLNULL

1+ 1

NULL

poot

!oid 'ain(!oid)

{

bool bnd;


%

ptr)<

true true

true

true

true

h f h


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Big Oh of BST SearchHuestion-

In the average BST ith 4 values&ho many steps are re?uire" to

fin" our valueG

Huestion-In the orst case BST ith4 values& ho many steps are

re?uire" fin" our valueG

Huestion-If there are + .illion no"es in a BST& ho

many steps ill it taKe to perform a searchG %O%,4o that's IN,

FQ eliminate",FQeliminate",

FQeliminate",

FQ

eliminate",

ight, log2346 steps

ight, 4 steps

Just *2,

I ti 4 l I t BST


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Inserting 4e alue Into BST

To insert a ne no"e in our BST& e must place the

ne no"e so that the resulting tree is still a vali"BST ,

%here oul" the folloingne values goG

@arly

Carly

Pen

Ken

lice

Alice

I ti 4 l I t BST


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If the tree is empty llocate a ne no"e an" put into it oint the root pointer to our ne no"e( $O4E,

Input- value to insert

Start at the root of the tree

%hile e're not "one

If is greater than current no"e's value If there is a right chil"& then go right ELSE allocate a ne no"e an" put into it&

set current no"e's right pointer to ne no"e( $O4E,

If is e?ual to current no"e's value& $O4E, 3nothing to "o(((6

If is less than current no"e's value If there is a left chil"& then go left ELSE allocate a ne no"e an" put into it& an" set current no"e's left pointer to ne no"e( $O4E,

4 h @ @ " ,

J t ith l

n" here's our Binary Search

void insert(const std::string &value)


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4o the @ @o"e,

struct 4o"e7

st"--string value: 4o"e 5left&5right:=:

Just as ith a regular.inary tree& e use a no"estruct to hol" our items(>oever let's a"" a constructor to

our 4o"e so e can easily create ane one,

4o"e3const st"--string Rmyal6 7 value 8 myal: left 8 right 8 4ULL:

=

class BinarySearchTree7pu.lic-

BinarySearchTree36 7 mroot 8 4ULL:

= voi" insert3const st"--string Rvalue6 7 =

private-

4o"e 5mroot:=:

Our BST class has a single

mem.er varia.le the rootpointer to the tree(

n" our constructor initialies thatroot pointer to 4ULL

hen e create a ne tree(3This in"icates the tree is empty6

n" here s our Binary SearchTree class(

4o let's see our complete

insertion function in @(

void insert(const std string &value) {

if (mroot !! NULL) { mroot ! ne" Node(value)# return# $

Node %cur ! mroot# for (##) {

if (value !! cur'value) return#

if (value cur'value) {

if (cur'left ! NULL)cur ! cur'left#

else { cur'left ! ne" Node(value)# return# $ $

else if (value ' cur'value) {

if (cur'rig*t ! NULL)cur ! cur'rig*t#

else {

cur'rig*t ! ne" Node(value)# return# $ $ $

$

If our tree isempty&

allocate ane no"e an"point the

root pointerto it then

e're "one,

Start traversing"on from theroot of the tree(

for3::6 is thesame as an

infinite loop(

If our value isalrea"y in the

tree& then e're"one ) Vust

return(

If the value toinsert is less

than the current

no"e's value&then go left(

If there is ano"e to our left&a"vance to that

no"e an"continue(

Otherise e'vefoun" the

proper spot forour ne value,

"" our value as

the left chil" ofthe current

no"e(

If the value e

ant to insert isgreater than thecurrent no"e's

value& thentraverse9insert

to the right(



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( g ) {






else { cur'left ! ne" Node(value)# return# $ $



else {


$

!oid 'ain(!oid)

{

inarSearch.ree bst;

bstinsert(0Larr1);

bstinsert(02hil1);

%

mroot 4ULL

/Larry0NULL NULL



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( g ) {






else {

cur'left ! ne" Node(value)# return# $ $



else {


$

!oid 'ain(!oid)

{

inarSearch.ree bst;

bstinsert(0Larr1);

bstinsert(02hil1);

%

mroot

NULL

/Larry0

/!ran0 /on"a0

NULL NULL

NULL

/Barry0

hil 88 LarryGGhil ; LarryGG

hil < LarryGG

Ghil 88 on"aGGhil ; on"aGG

G

/hil0NULL NULL

cur

I i 4 l I BST


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s ith BST Search& there is a recursive version ofthe Insertion algorithm too( Be familiar ith it,

Huestion-

Miven a ran"om array of num.ers if you insert them oneat a time into a BST& hat ill the BST looK liKeG

Huestion-

Miven a or"ere" array of num.ers if you insert themone at a time into a BST& hat ill the BST looK liKeG

Bi Oh f BST I ti


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Big Oh of BST Insertion

So& hat's the .ig)oh of BST InsertionGight, It's also O3log2n6

%hyG Because e have to first use a .inary search to fin"

here to insert our no"e an" .inary search is O3log2n6(

Once e've foun" the right spot& e can insert our neno"e in O316 time(

Mroovy Ba.y,

!i "i Ni R N f BST


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Huestion- %hat's the .ig)oh to fin" the minimum or

maximum elementG

NULL

/Larry0

/!ran0 /on"a0

NULL NULL

NULL

/Barry0 /hil0NULL NULL

!in"ing Nin R Nax of a BST >o "o e fin" the minimum an" maximum values in a BSTG

int 3et4in(node *p+oot)

{ if (p+oot == NULL)

return(-); // e'pt

&hile (p+oot-left "= NULL)

p+oot = p+oot-left;

return(p+oot-!alue);

%

int 3et4a5(node *p+oot)

{ if (p+oot == NULL)

return(-); // e'pt

&hile (p+oot-ri$ht "= NULL)

p+oot = p+oot-ri$ht;


%

The minimum value is locate" at the left)most no"e( The maximum value is locate" at the right)most no"e(

!i "i Ni R N f BST


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>opefully you're getting the i"ea that most treefunctions can .e "one recursively

!in"ing Nin R Nax of a BST n" here are recursive versions for you

int 3et4in(node *p+oot)

{

if (p+oot == NULL)

return(-); // e'pt

if (p+oot-left == NULL)


return(3et4in(p+oot-left));

%

int 3et4a5(node *p+oot)

{

if (p+oot == NULL)

return(-); // e'pt

if (p+oot-ri$ht == NULL)


return(3et4a5(p+oot-ri$ht));

%

i i BST I l h . i l O "


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InOr"er3cur)


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!reeing The %hole Tree

%hen e are "one ith our BST& e have to free every

no"e in the tree& one at a time(

Huestion- @an anyone thinK of an algorithm for thisG

voi" !reeTree34o"e 5cur67 if 3cur 88 4ULL6 99 if empty& return return:

!reeTree3cur)


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voi" !reeTree34o"e 5cur67 if 3cur 88 4ULL6

return: !reeTree3cur)


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! i Th %h l T


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NULL

/Larry0

/!ran0 /on"a0NULL

/Ma..y0NULL NULL

cur)<

cur)<




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NULL

/Larry0

/!ran0 /on"a0NULL

cur)<

cur)<




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Lecture12 Updated