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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.

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2 Answers 2

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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;
    }
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  • \$\begingroup\$ 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. \$\endgroup\$ Commented Feb 11, 2022 at 12:19
  • \$\begingroup\$ @Imus Defenetly a bug in OP's code, as you need to have at least 1 node to be able to call size(). \$\endgroup\$ Commented Feb 11, 2022 at 12:43
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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;
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  • \$\begingroup\$ I have literally no experience with Ternary Operators, so I really have no idea what that code is meant to mean \$\endgroup\$ Commented Feb 3, 2022 at 16:30

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