Counting nodes in a tree in Java

Question

First of all, I swear this is not homework, it's a question I was asked in an interview. I think I made a mess of it (though I did realise the solution requires recursion). Here is the question:

Implement the count() method which returns the number of nodes in a tree. If a node doesn't have either a left or right child, the relevant getXXChild() method will return null

class Tree {

  Tree getRightChild() {
    // Assume this is already implemented
  }

  Tree getLeftChild() {
    // Assume this is already implemented
  }

  int count() {
    // Implement me
  }
}

My reason for asking the question is simply curious to see the correct solution, and thereby measure how bad mine was.

Cheers, Tony

Answer 1

My first attempt didn't have anything new to add, but then I started to wonder about recursion depth and whether it would be possible to rearrange the code to take advantage of the tail call optimization feature of the latest Java compiler. The main problem was the null test - which can be solved using a NullObject. I'm not sure if TCO can deal with both recursive calls, but it should at least optimize the last one.

static class NullNode extends Tree {

    private static final Tree s_instance = new NullNode();

    static Tree instance() {
        return s_instance;
    }

    @Override
    Tree getRightChild() {  
        return null;
    }  

    @Override
    Tree getLeftChild() {  
        return null;
    }  

    int count() {  
        return 0;
    }
}

int count() {      
    Tree right = getRightChild();      
    Tree left  = getLeftChild();      

    if ( right == null ) { right = NullNode.instance(); }
    if ( left  == null ) { left  = NullNode.instance(); }

    return 1 + right.count() + left.count();
}   

The precise implementation of NullNode depends on the implementations used in Tree - if Tree uses NullNode instead of null, then perhaps the child access methods should throw NullPointerException instead of returning null. Anyway, the main idea is to use a NullObject in order to try to benifit from TCO.

Answer 2

Questions related to binary tree should be expected in an interview. I would say to take time before any next interview and go through this link. There are about 14 problems solved .You can have a look and how the solution is done. This would give you an idea of how to tackle a problem with binary tree in future.

I know your question is specific to the count method .That is also implemented in the link that i provided

Answer 3

Something like this should work:

int count()
{
    int left = getLeftChild() == null ? 0 : getLeftChild().count();
    int right = getRightChild() == null ? 0 : getRightCHild().count();

    return left + right + 1;
}

Answer 4

Implement the method:

public static int countOneChild(Node root)
{
    ...
}

that counts the number of internal nodes in a binary tree having one child. Add the function to tree.java program.

Answer 5

int count()

{
   int retval = 1;
    if(null != getRightChild()) retval+=getRightChild().count();
    if(null != getLeftChild()) retval+=getLeftChild().count();
    return retval;

}

God I hope I didn't make a mistake.

EDIT: I did actually.

Answer 6

I like this better because it reads:

return count for left + count for rigth + 1

  int count() {
      return  countFor( getLeftChild() ) + countFor( getRightChild() ) + 1;
  }
  private int countFor( Tree tree )  { 
       return tree == null ? 0 : tree.count();
  }

A little more towards literate programming.

BTW, I don't like the getter/setter convention that is so commonly used on Java, I think a using leftChild() instead would be better:

  return countFor( leftChild() ) + countFor( rightChild() ) + 1;

Just like Hoshua Bloch explains here http://www.youtube.com/watch?v=aAb7hSCtvGw at min. 32:03

If you get it rigth your code reads...

BUT, I have to admit the get/set convention is now almost part of the language. :)

For many other parts, following this strategy creates self documenting code, which is something good.

Tony: I wonder, what was your answer in the interview.