number of nodes for a level in a first child - next sibling tree

Question

I have to write a function to find the number of nodes in a level of a first child - next sibling N-ary tree. My function is:

int nodesAtLevel(NTree root, int level) {
    if (root == NULL) {
        return 0;
    }
    if (level == 0) {
        return 1;
    }
    return nodesAtLevel(root->firstChild, level - 1) + 
           nodesAtLevel(root->nextSibling, level - 1);
}

but it does not work. Can someone help me?

*but it does not work* How should it work?How is it not working? — Gaurav Sehgal, Jul 27 '17 at 09:47
Can you explain why you are decrementing `level` when calculating the number of nodes associated with `nextSibling`? — r3mainer, Jul 27 '17 at 09:48
The structure of your tree is unclear. This is an n-ary tree? So not only does a node have children it has siblings? And if that's the case, then `if(level == 0) { return 1; }` wouldn't be correct. What if the current level has siblings? The answer would not then be 1. If you haven't done so, draw yourself a picture of the tree structure and use it as a guide to understand your code. — lurker, Jul 27 '17 at 10:24

score 0 · Answer 1 · edited Aug 10 '17 at 07:07

Right now, your code seems to only return 2. I believe this is what you are trying to do:

int nodesAtLevel(NTree root, int level) {
    if (root == NULL) {
        return 0;
    }
    if (level == 0) {
        return 1;
    }

    int x = nodesAtLevel(root->firstChild, level - 1);
    int y = nodesAtLevel(root->nextSibling, level - 1);

    return x + y + 1; //add 1 for current node
}

This should update the value after every recursion, unlike your current code.

number of nodes for a level in a first child - next sibling tree

1 Answers1