Golgappa.net | Golgappa.org | BagIndia.net | BodyIndia.Com | CabIndia.net | CarsBikes.net | CarsBikes.org | CashIndia.net | ConsumerIndia.net | CookingIndia.net | DataIndia.net | DealIndia.net | EmailIndia.net | FirstTablet.com | FirstTourist.com | ForsaleIndia.net | IndiaBody.Com | IndiaCab.net | IndiaCash.net | IndiaModel.net | KidForum.net | OfficeIndia.net | PaysIndia.com | RestaurantIndia.net | RestaurantsIndia.net | SaleForum.net | SellForum.net | SoldIndia.com | StarIndia.net | TomatoCab.com | TomatoCabs.com | TownIndia.com
Interested to Buy Any Domain ? << Click Here >> for more details...


Write a program to get the binary tree.

Answers were Sorted based on User's Feedback



Write a program to get the binary tree...

Answer / r.r.bharti

Simple way:

public int FindDepthOfTree(RBNode n)
{
if (n == null) return 0;
return Math.Max(FindDepthOfTree(n.LeftNode),
FindDepthOfTree(n.RightNode)) + 1;
}

Is This Answer Correct ?    1 Yes 0 No

Write a program to get the binary tree...

Answer / vamshi koduri

#include<iostream>
using namespace std;
typedef struct node
{
int d;
struct node *l,*r;
}node;
class bt
{
public:
int h,lev;
node *root;
node *create();
int height(node *);
void printlevel(node *,int lev);
void givenlevel();
void inorder(node *);
void preorder(node *);
void postorder(node *);
}a;
node *bt::create()
{
struct node *n;
n=new node;
int x;
cout<<"enter element
";
cin>>x;
if(x==-1)
return NULL;
n->d=x;
n->l=n->r=NULL;
cout<<"enter left "<<n->d<<ends;
n->l=create();
cout<<"enter right "<<n->d<<ends;
n->r=create();
return n;
}
void bt::inorder(node *p)
{
if(p!=NULL)
{
inorder(p->l);
cout<<p->d<<ends;
inorder(p->r);
}
}
void bt::preorder(node *q)
{
if(q!=NULL)
{
cout<<q->d<<ends;
preorder(q->l);
preorder(q->r);
}
}
void bt::postorder(node *q)
{
if(q!=NULL)
{
postorder(q->l);
postorder(q->r);
cout<<q->d<<ends;
}
}
int main()
{
a.root=a.create();
cout<<"levelorder: ";
a.givenlevel();
cout<<endl;
cout<<"inorder : ";
a.inorder(a.root);
cout<<endl;
cout<<"preorder : ";
a.preorder(a.root);
cout<<endl;
cout<<"postorder : ";
a.postorder(a.root);
cout<<endl<<"height: ";
cout<<a.height(a.root);
return 0;
}
int bt::height(node *p)
{
if(p==NULL)
return 0;
int l=height(p->l);
int r=height(p->r);
if(l>r)
h=1+l;
else
h=1+r;
return h;
}
void bt::givenlevel()
{
int p=height(root);
for(int i=1;i<=h;i++)
printlevel(root,i);
}
void bt::printlevel(node *p,int lev)
{
if(p==NULL)
return;
if(lev==1)
cout<<p->d<<ends;
if(lev>1)
{
printlevel(p->l,lev-1);
printlevel(p->r,lev-1);
}
}

Is This Answer Correct ?    0 Yes 0 No

Write a program to get the binary tree...

Answer / vijayan

#include<iostream.h>

Is This Answer Correct ?    4 Yes 21 No

Post New Answer

More OOPS Interview Questions

What is constructor overloading in oop?

0 Answers  


In multiple inheritance , to create sub class object , is there need to create objects for its superclasses??? in java and c++ both. Actually i have some information that is , all available members from its superclasses , memory created in subclass obj , so no need to create object for its superclasses...??? Thanks in Advance

1 Answers  


Child cObj = new Parent() Wahts the output ?

8 Answers   Patni, TCS,


• What are the desirable attributes for memory managment?

0 Answers  


why we call c++ is object oriented lanaguage

7 Answers   HCL,


which is best institute to learn c,c++ in ameerpet hyderabad

1 Answers  


What is encapsulation oop?

0 Answers  


How to calculate the age from the date of birth by using the program?

2 Answers   Accenture,


what is object oriented programming and procedure oriented programming?

3 Answers  


Are polymorphisms mutations?

0 Answers  


when my application exe is running nad i don't want to create another exe what should i do

2 Answers   HCL,


suppose A is a base class and B is the derved class. Both have a method foo which is defined as a virtual method in the base class. You have a pointer of classs B and you typecast it to A. Now when you call pointer->foo, which method gets called? The next part of the question is, how does the compiler know which method to call?

3 Answers   EA Electronic Arts,


Categories