C++ PROGRAM FOR BINARY TREE TRAVERSAL

BINARY TREE IMPLEMENTATION AND TRAVERSAL

#include<iostream.h>
#include<alloc.h>
#include<stdlib.h>
#include<conio.h>
struct freeptr
{
unsigned entry;
struct freeptr *right,*left;
}*root;
char waste;
void preorder(struct freeptr *root)
{
if(root)
{
cout<<"\t"<<root->entry;
preorder(root->left);
preorder(root->right);
}
}
void inorder(struct freeptr *root)
{
if(root)
{
inorder(root->left);
cout<<"\t"<<root->entry;
inorder(root->right);
}
}
void postorder(struct freeptr *root)
{
if(root)
{
postorder(root->left);
postorder(root->right);
cout<<"\t"<<root->entry;
}
}
int init(struct freeptr *t)
{
cin>>t->entry;
if(t->entry)
{
t->left=(struct freeptr*)malloc(sizeof(struct freeptr));
cout<<"the lest child of:"<<t->entry;
cout<<"\t";
if(init(t->left)==0)
{
free(t->left);
t->left=NULL;
}
cout<<"the right child of:"<<t->entry;
cout<<"\t";
t->right=(struct freeptr*)malloc(sizeof(struct freeptr));
if(init(t->right)==0)
{
free(t->right);
t->right=NULL;
}
return 1;
}
else
return 0;
}
void main()
{
clrscr();
root=(struct freeptr*)malloc(sizeof(struct freeptr));
cout<<"\n BINARY TREE TRAVERSAL-USING RECURSION"<<endl;
cout<<"enter the binary tree starting from the root node"<<endl;
init(root);
cout<<"press any key.."<<endl;
waste=getch();
cout<<"by preorder traversal"<<endl;
preorder(root);
cout<<endl;
cout<<"by inorder traversal"<<endl;
inorder(root);
cout<<endl;
cout<<"by postorder traversal"<<endl;
postorder(root);
cout<<endl;
getch();
}

OUTPUT:

 BINARY TREE TRAVERSAL-USING RECURSION

enter the binary tree starting from the root node
1
the lest child of:1     5
the lest child of:5     1
the lest child of:1     2
the lest child of:2     3
the lest child of:3     0
the right child of:3    0
the right child of:2    1
the lest child of:1     2
the lest child of:2     5
the lest child of:5     0
the right child of:5    0
the right child of:2    0
the right child of:1    0
the right child of:1    0
the right child of:5    0
the right child of:1    0
press any key..
by preorder traversal
        1       5       1       2       3       1       2       5
by inorder traversal
        3       2       5       2       1       1       5       1
by postorder traversal
        3       5       2       1       2       1       5       1