Binary Search Tree complete operations

#include<stdio.h>
#include<conio.h>
#include<malloc.h>
struct node
{
    int data;
    struct node *l,*r;
};
struct node *root=NULL;
int item,item2,item3;
struct node *insert(struct node *root,int item);
void search(struct node *root,int item);
void preorder(struct node *root);
void inorder(struct node *root);
void postorder(struct node *root);
void delnode(struct node *root,int item3);

void main()
{
    int choice,ch=0;
    do
    {
        printf("\t\t\tBinary Search Tree Operations :\n");
        printf("\t\t1.Insert a node in the tree.\n");
        printf("\t\t2.Traverse and display the tree in inorder.\n");
        printf("\t\t3.Traverse and display the tree in preorder.\n");
        printf("\t\t4.Traverse and display the tree in postorder.\n");
        printf("\t\t5.Delete a node from the tree.\n");
        printf("\t\t6.Search for an element.\n");
        printf("Enter your choice : ");
        scanf_s("%d",&choice);
        switch(choice)
        {
        case 1:root=insert(root,item);
            break;
        case 2:inorder(root);
            break;
        case 3:preorder(root);
            break;
        case 4:postorder(root);
            break;
        case 5:delnode(root,item3);
            break;
        case 6:search(root,item2);
            break;
        default:printf("Wrong choice !!\n");
            break;
        }
        printf("Press 0 to continue.... ");
        scanf("%d",&ch);
    }while(ch==0);
}

struct node *insert(struct node *root,int item)
{
    struct node *p,*f;
    printf("Enter the item : ");
    scanf_s("%d",&item);
    p=root;
    if(root==NULL)
    {
        root=(struct node *)malloc(sizeof(struct node));
        root->data=item;
        root->l=NULL;
        root->r=NULL;
       
    }
    else
    {
       
        while(p!=NULL)
        {
            f=p;
            if(item>p->data)
            {
                p=p->r;
            }
            else
                p=p->l;
        }
        if(item>f->data)
        {
            f->r=(struct node *)malloc(sizeof(struct node));
            f=f->r;
            f->data=item;
            f->l=f->r=NULL;
        }

        else
        {
            f->l=(struct node *)malloc(sizeof(struct node));
            f=f->l;
            f->data=item;
            f->l=f->r=NULL;
        }
    }
    return root;
}

void inorder(struct node *root)
{
    struct node *p;
    p=root;
    if(p!=NULL)
    {
       
       
        inorder(p->l);
        printf("%3d",p->data);
        inorder(p->r);
    }
}

void preorder(struct node *root)
{
    struct node *p;
    p=root;
    if(p!=NULL)
    {
       
       
        printf("%d",p->data);
        preorder(p->l);
        preorder(p->r);
    }
}

void postorder(struct node *root)
{
    struct node *p;
    p=root;
    if(p!=NULL)
    {
       
       
        postorder(p->l);
        postorder(p->r);
        printf("%d",p->data);
    }
}

void search(struct node *root,int item2)
{
    struct node *p;
    if(root==NULL)
    {
        printf("Tree is empty!!\n");
        return;
    }
    else
    {
        p=root;
        printf("Enter the element to be searched : ");
        scanf("%d",&item2);
        while(p!=NULL)
        {
            if(item2==p->data)
            {
                printf("Item found.\n");
                return;
            }
            else if(item2>p->data)
                p=p->r;
            else
                p=p->l;
        }
        printf("Element not found.\n");
    }
}

void delnode(struct node *root,int item3)
{
    struct node *curr,*par,*temp,*tmp;
    int found=0; curr=root;
    if(curr==NULL)
    {
        printf("The tree is empty !!\n");
        return;
    }
    else
    {
        printf("Enter the element to delete : ");
        scanf("%d",&item3);
        while(curr!=NULL)
        {
            par=curr;
            if(item3==curr->data)
            {
                found=1;
                break;
            }
            else if(item3<curr->data)
                curr=curr->l;
            else
                curr=curr->r;
        }
        if(!found)
        {
            printf("Element not found !!\n");
            return;
        }
        if(curr->l==NULL && curr->r==NULL) //Leaf node
        {
            if(par->l==curr)
                par->l=NULL;
            else
                par->r=NULL;
            free(curr);
            return;
        }
        //Node having only RST
        if(curr->l==NULL && curr->r!=NULL) 
        {
            if(par->l==curr)
            {
                par->l=curr->r;
                free(curr);
            }
            else
            {
                par->r=curr->r;
                free(curr);
            }
            return;
        }
        //Node having only LST
        if(curr->l!=NULL && curr->r==NULL)
        {
            if(par->l==curr)
            {
                par->l=curr->l;
                free(curr);
            }
            else
            {
                par->r=curr->l;
                free(curr);
            }
            return;
        }
        if(curr->l!=NULL && curr->r!=NULL)
        {
            //Replace current node's data with smallest element
            struct node *chkr,*lcurr,*lcurrp,*tmp;
            chkr=curr->r;
            if((chkr->l==NULL) && (chkr->r==NULL))
            {
                curr->data=chkr->data;
                curr->r=NULL;
                free(chkr);
                return;
            }
            else if(chkr->l!=NULL)
            {
                //lcurr is the node to be replaced
                lcurr=chkr;
                //lcurrp is the parent of lcurr
                lcurrp=chkr->l;
                while(lcurr->l!=NULL)
                {
                    lcurrp=lcurr;
                    lcurr=lcurr->l;
                }
                curr->data=lcurr->data;
                lcurrp->l=NULL;
                free(lcurr);
            }
            else
            {
                tmp=curr->r;
                curr->data=tmp->data;
                curr->r=tmp->r;
                free(tmp);
            }
        }
   }
   printf("Element terminated\n");
}

       

Can u guess the output... U jst need a simple concept of bitwise and logical operators in C

#include<stdio.h>
int main(void)
{
    int a[10],i=0,s=0;
    printf("Enter 5 digits (choose digits between 0-9)::\n");
    for(i=0;i<5;i++)
        scanf("%d",&a[i]);


    printf("\n\n");
    for(i=0;i<4;i++)
        printf("%5d|%d=%d",a[i+1],a[i],a[i+1]|a[i]);



    printf("\n\n");
    for(i=0;i<4;i++)
        printf("%5d||%d=%d",a[i+1],a[i],a[i+1]||a[i]);
  
  

Some interesting questions in C u would have never guessed :)

Predict the output or error(s) for the following:

1.    void main()
{
    int  const * p=5;
    printf("%d",++(*p));
}
Answer:
        Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the
value of the "constant integer".

2.    main()
{
    char s[ ]="man";
    int i;
    for(i=0;s[ i ];i++)
    printf("%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
                mmmmaaaannnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the
same idea. Generally  array name is the base address for that array. Here s is
the base address. i is the index number/displacement from the base
address. So, indirecting it with * is same as s[i]. i[s] may be
surprising. But in the  case of  C  it is same as s[i].

3.    main()
{
    float me = 1.1;
    double you = 1.1;
    if(me==you)
printf("I love U");
else
        printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values
cannot
be predicted exactly. Depending on the number of bytes, the precession
with of the value  represented varies. Float takes 4 bytes and long
double
takes 10 bytes. So float stores 0.9 with less precision than long
double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers
with relational operators (== , >, <, <=, >=,!= ) .

4.    main()
    {
    static int var = 5;
    printf("%d ",var--);
    if(var)
        main();
    }
Answer:
5 4 3 2 1
            Explanation:
When static storage class is given, it is initialized once. The change
in the value of a static variable is retained even between the function
calls. Main is also treated like any other ordinary function, which can
be called recursively.

5.    main()
{
     int c[ ]={2.8,3.4,4,6.7,5};
     int j,*p=c,*q=c;
     for(j=0;j<5;j++)
         {
        printf(" %d ",*c);
           ++q;     
         }
     for(j=0;j<5;j++)
         {
           printf(" %d ",*p);
           ++p;     
         }
}

Answer:
                2 2 2 2 2 2 3 4 6 5
             Explanation:
Initially pointer c is assigned to both p and q. In the first loop,
since only q is incremented and not c , the value 2 will be printed 5 times.
In second loop p itself is incremented. So the values 2 3 4 6 5 will be
printed.

6.    main()
{
    extern int i;
    i=20;
        printf("%d",i);
}

Answer:
Linker Error : Undefined symbol '_i'
Explanation:
                 extern storage class in the following declaration,
                               extern int i;
specifies to the compiler that the memory for i is allocated in some
other program and that address will be given to the current program at the
time of linking. But linker finds that no other variable of name i is
available in any other program with memory space allocated for it. Hence a linker
error has occurred .

7.    main()
{
    int i=-1,j=-1,k=0,l=2,m;
    m=i++&&j++&&k++||l++;
    printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
                0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the
logical AND (&&) operator has higher priority over the logical OR (||)
operator.So the expression  ‘i++ && j++ && k++’ is executed first. The result of
this expression is 0(-1 && -1 && 0 = 0). Now the expression is 0 || 2
which evaluates to 1 (because OR operator always gives 1 except for
‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The
values of other variables are also incremented by 1.

8.    main()
{
    char *p;
    printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
                1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P
is a character pointer, which needs one byte for storing its value (a
character). Hence sizeof(*p) gives a value of 1. Since it needs two
bytes to store the address of the character pointer sizeof(p) gives 2.

9.    main()
{
    int i=3;
    switch(i)
     {
        default:printf("zero");
        case 1: printf("one");
           break;
       case 2:printf("two");
          break;
      case 3: printf("three");
          break;
      }
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed
only when all other cases doesn't match.

10.    main()
{
      printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times
the least significant 4 bits are filled with 0's.The %x format specifier
specifies that the integer value be printed as a hexadecimal value.

11.    main()
{
    char string[]="Hello World";
    display(string);
}
void display(char *string)
{
    printf("%s",string);
}
              Answer:
Compiler Error : Type mismatch in redeclaration of function display
              Explanation :
In third line, when the function display is encountered, the compiler
doesn't know anything about the function display. It assumes the
arguments and return types to be integers, (which is the default type). When it
sees the actual function display, the arguments and type contradicts with
what it has assumed previously. Hence a compile time error occurs.

12.    main()
{
    int c=- -2;
    printf("c=%d",c);
}
Answer:
                     c=2;
              Explanation:
Here unary minus (or negation) operator is used twice. Same maths 
rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can  only be
applied to variables as a decrement operator (eg., i--). 2 is a constant and
not a variable.

13.    #define int char
main()
{
    int i=65;
    printf("sizeof(i)=%d",sizeof(i));
}
Answer:
                  sizeof(i)=1
Explanation:
Since the #define replaces the string  int by the macro char

14.    main()
{
int i=10;
i=!i>14;
printf("i=%d",i);
}
Answer:
i=0


     Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than
‘>’symbol.  ! is a unary logical operator. !i (!10) is 0 (not of true is
false).  0>14 is false (zero).

15.    #include<stdio.h>
main()
{
char s[]={'a','b','c','
','c','

Warshall's Algorithm

#include<stdio.h>
#include<conio.h>
void main()
{
    int a[10][10],i,j,k,n;
    printf("Enter the no. of nodes : ");
    scanf("%d",&n);
    //Enter only 1 or 0
    printf("Now enter the matrix row wise : \n");
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            scanf("%d",&a[i][j]);
    printf("\nThe matrix u entered is : \n");
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
            printf("%2d",a[i][j]);
        printf("\n");
    }

    //To find out the transitive relation matrix
    for(k=0;k<n;k++)
    {
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                a[i][j]=a[i][j]||(a[i][k]&&a[k][j]);
    }

    //To print the required transitive relation matrix
    printf("\n\nThe required transitive matrix is : \n");
    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
            printf("%2d",a[i][j]);
        printf("\n");
    }
}

N.B. : Check this program for 5x5 matrix and lemme know if there is any fault...