size() method for a binary search tree

Question

This is my size method for my binary search tree, that is meant to implement recursion to calculate the tree's size.

 public int size() { if (left != null && right != null) { return left.size() + 1 + right.size(); } else if (left != null && right == null) { return left.size() + 1; } else if (right != null && left == null) { return right.size() + 1; } else { return 0; } } 

First I'm wondering if this looks all right. I also got some feedback on this function that I can calculate the size of the tree with fewer if statements but I don't see how I can do that.

Answer 1

Can be made much simpler:

public int size() { int size=1; if(left != null) size+=left.size(); if(right != null) size+=right.size(); return size; } 

Answer 2

Seems strange for the else case to return zero. I would expect it to return 1 (the node itself).

Then it becomes much simpler:

int leftSize = left == null ? 0 : left.size(); int rightSize = right == null ? 0 : right.size(); return 1 + leftSize + rightSize;