function to print whole path of nodes

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3,1 //connected to each other directed 3 to 1
3,2
3,5
1,6
1,7
5,8

//function used to print path
void descendents(struct node * n)
{
if (n!=NULL)
{printf("-> %d",n->data);
descendents(n->next1);
descendents(n->next2);
descendents(n->next3);
}
}

for 3
output is

3->1->6->7->2->5->8

but i want output such that

3->1->6
3->1->7
3->2
3->5->8

//what changes in function i have to do

Posted 22-Aug-14 5:42am

Member 11018648

Updated 22-Aug-14 6:19am

v3

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Sergey Alexandrovich Kryukov 22-Aug-14 12:14pm

If you start your sentence from the middle, as you do, don't expect people to understand it.
—SA

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KarstenK · Answer 1 · 2014-08-22T06:42:00

Solution 1

you must redesign the interface

C++

int descendents(struct node * n, int index = 0);//print only the n-node path

int r = 0;

while( r > -1 )
{
  r = descendents( n, r );
}

And implement it.
It will be a bit tricky, because you must recurse somehow BUT remember where you have been.

Posted 22-Aug-14 6:42am

KarstenK

nv3 · Answer 2 · 2014-08-22T07:16:00

You are traversing the tree in so-called in-order sequence. That's okay. To achieve what you want, you should do the following:

(a) store in every node its parent node (in addition to the links you already have)

(b) print only something when you have reached an leaf-node (one without further down-links)

(c) in that case print the entire path from the root to your node, which is a little tricky:

(c1) store the nodes from your leaf-node upward by backtracking along the parent links to the root; store them in an array or stack

(c2) print the stored nodes in reverse order, i.e. from root to leaf

Hope that helps.