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

return (getRightChild() == null ? 0 : getRightChild.count()) + (getLeftChild() == null ? 0 : getLeftChild.count()) + 1;

Or something like that.

Answer 2

You can count the tree by traversing it many ways. Simply preorder traversal, the code would be (based on the functions you defined):

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

Answer 3

class Tree {

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

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

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

Answer 4

class Tree {

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

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

 int count() {
    if(this.getLeftChild() !=null && this.getRightChild()!=null) 
        return 1 + this.getLeftChild().count() + this.getRightChild().count();
    elseif(this.getLeftChild() !=null && this.getRightChild()==null)
        return 1 + this.getLeftChild().count();
    elseif(this.getLeftChild() ==null && this.getRightChild()!=null)
        return 1 + this.getRightChild().count();
    else return 1;//left & right sub trees are null ==> count the root node
  }
}

Answer 5

I did it by preorder recurssion. Altough it doesn't exactly follow the interview format by using localRoot, but I think you get the idea.

private int countNodes(Node<E> localRoot, int count) {
    if (localRoot == null) 
        return count;     
    count++; // Visit root
    count = countNodes(localRoot.left, count); // Preorder-traverse (left)
    count = countNodes(localRoot.right, count); // Preorder-traverse (right)
    return count;
}

public int countNodes() {
   return countNodes(root, 0);
}

Answer 6

This is a standard recursion problem:

count():
    cnt = 1 // this node
    if (haveRight) cnt += right.count
    if (haveLeft)  cnt += left.count
return cnt;

Very inefficient, and a killer if the tree is very deep, but that's recursion for ya...