## basukinjal Newbie Poster

Suppose we have two array inorder and preorder containing the elements in the said format.
Using these two arrays how can i generate the binary tree??

Any help on the algorithm to be followed will be highly appreciated.

For eg:
inorder is : 2 3 4 5 6 7 8
preorder is : 5 3 2 4 7 6 8

then the Binary tree shld be
____5
____/\
_3_7
___/_/\
2_4 6_8

How can i form this binary tree??
i know the algorithm by hand but how do i implement it in C??

## basukinjal Newbie Poster

Ive made my binary tree in the following format
[code=C]
typedef struct node
{ int info;
struct node left;
struct node
right;
}*nodeptr;
[/code]

Im looking for a function that will accept the root pointer and print the tree in a graphical format.
for eg.
i want the output to be
5
/\
__37
__//\
_268
____
\
__9
How can i get such and output (without the underscores .. given only because spaces cannot be detected.)

## basukinjal Newbie Poster

I have a broadband connection that i want to dial at a specific time automatically.
Can anyone tell me how to do it??

## basukinjal Newbie Poster

Ive written the following simple code to find the addresses of the varialbles

[code=C]

include

int main()
{ int a;
int b;
char c;
float d;
printf("\nInt : %x\nInt : %x\nChar : %x\nfloat : %x",&a,&b,&c,&d);
return 0;
}

/* OUTPUT

Int : 12ff4c
Int : 12ff48
Char : 12ff47
float : 12ff40

*/
[/code]

My Question is why is there an extra gap of 3 in the memory??

## basukinjal

Lets say i hv the permutation
2,5,4,1,0,7,6,3.

Now i want to bring it to
0,1,2,3,4,5,6,7.
Sorting will easily do the thing for me.

But im looking for a better algorithm than sorting.
i.e. since sorting (any type) will bring nos say 1,9,6,7,0 to 0,1,6,7,9
It doesn't take into consideration that all nos from 0 to 9 are present or not.

But i know that all nos are from 0 to N -1 (N = 8 in given case) are present in the array.
So can i can i modify the sorting algorithm in any way such that the time complexity is reduced? i.e a better algorithm with the known given condition.

## basukinjal

I know what identity mapping is.
i.e. Array[i]=i.
From the random permutation i can come to this stage by sorting.
My question if can i do any better?? i.e sorting algorithm does not take into fact that all nos are present. but since i already know that all nos are present..can i do anything better to solve the problem. Im looking for a better algorithm than sorting.

## basukinjal Newbie Poster

Suppose i have a random permutation of 0,1,...., N-1. Now if i want to get the identity mapping from this can i do anything better than sorting?? I mean, sorting does not take into fact that all nos from 0 to N-1 are in the array or not. But in this case as we know that all the nos from 0 to N-1 are present...can I get a better algorithm and just sorting the array??