Filtering inorder bst print traversal

Hi,
Does anyone have an idea of how to filter an inorder print traversal for a bst?
Let's say u have 2 + nodes carrying the same data, and u only wanna print one of it out.

this is my problem.

my print output seem to be like (eg) - [number is the frequency]

A 4
A 4
A 4
AA 2
AA 2
GAA 2

and i want it to be

A 4
AA 2
GAA 2

.... I thought of something like,

Code:

void print_inorder(node_t *p_ptr, char *s){
     node_t *x, *y;
     
     x = p_ptr->left;     
     y = p_ptr->right;
     
	 if(p_ptr!=NULL){
        if(substring(s, p_ptr->start, p_ptr->end) == substring(s, x->start, x->end))
                     p_ptr=x;
        if(substring(s, p_ptr->start, p_ptr->end) == substring(s, y->start, y->end))
                     p_ptr=y;
		print_inorder(p_ptr->left,s);
		printf("%s %d\n", substring(s, p_ptr->start, p_ptr->end), p_ptr->freq);
		print_inorder(p_ptr->right,s);
	}
}

to skip over the node carrying the reoccuring data. But it ended up in a 'SEG FAULT', and seems crappy to me too...

Why do you have multiple nodes containing the same information?

oh yea, silly me

ok..i changed it to..

Code:

void print_inorder(node_t *p_ptr, char *s){
     
	 if(p_ptr!=NULL){
         if(substring(s, p_ptr->start, p_ptr->end) == substring(s, p_ptr->left->start, p_ptr->left->end))
                     p_ptr = p_ptr->left;
        if(substring(s, p_ptr->start, p_ptr->end) == substring(s, p_ptr->right->start, p_ptr->right->end))
                     p_ptr = p_ptr->right;
		print_inorder(p_ptr->left,s);
		printf("%s %d\n", substring(s, p_ptr->start, p_ptr->end), p_ptr->freq);
		print_inorder(p_ptr->right,s);
	}
}

but it still ends up in a seg fault..

Write a main() which just adds 3 nodes to your tree (one root, one left, one right) and then run it in the debugger.

If you can't figure it out from that, at least you will have a self-contained and complete program to post on the forum.

We can't tell what is possibly going wrong from one random snippet which may not even be the problem.

>Why do you have multiple nodes containing the same information

We can have two or more nodes containing the same data in bst right?
This is what a famous book has to say about this issue

" Let x be a node in a binary search tree. If y is a node in the left subtree of x, then key[y]
≤ key[x]. If y is a node in the right subtree of x, then key[x] ≤ key[y]. "

Though this only leads to some weird situations, it is theoritically possible .

>We can have two or more nodes containing the same data in bst right?
That's pointless, really. However, when you consider keys and values then it makes more sense. Different values can have the same keys and nodes are manipulated according to the key. In that case a tree can be designed to allow duplicates.

Hey guys!
I fixed the problem in another way already, thx to the suggestion about creating my own main func to test the print func.
Then it hit me...
I was so stupid!

I altered the tree not to store the reoccuring strings!! Y ?? y?? am i so dumb!
THX