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;
 }

• Imus: This does assume that if left and right are both null the size is also 1 where OP's code would return 0. Might have been a bug in OP's code.
• convert: @Imus Defenetly a bug in OP's code, as you need to have at least 1 node to be able to call 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;

• Rasmus Steen: I have literally no experience with Ternary Operators, so I really have no idea what that code is meant to mean