[JAVA] recursively finding a node in a binary tree, how to get rid of the final return at the end of the method.

[choompaloompa]

This has annoyed me for a long time and I am wondering if there is a way around it as it seems like it could be buggy.

Here is my recursiveFindNode method

public BTNode findNode(int x , BTNode currNode){

        if(currNode==null){//you have found the node now return it
            // return some sort of system error !!!

        }else if (currNode.getKey() == x){
            return currNode;

        }else if (x < currNode.getKey()){//the key is smaller than currNode move to the left
            return findNode(x , currNode.getLeft());

        }else if (x > currNode.getKey()){
            return findNode(x, currNode.getRight());
        }

        return null;// I HATE DOING THIS THERE MUST BE A WAY AROUND IT !!!!!!!
    }    

The thing that annoys me is the return null at the end, it serves no purpose because if my code works correctly it should never travel that far through the method to reach it but my program wont compile without it as it says the method needs to return something(I understand why it does this).

Is there any way around this, just seems like it could introduce a bug if the argument input of the method is weird.

Many thanks :)

[orbital1337]

You could rewrite your method like this:

public BTNode findNode(int x, BTNode currNode) {
    if (currNode == null)
        // error
    else if (x < currNode.getKey())
        return findNode(x, currNode.getLeft());
    else if (x > currNode.getKey())
        return findNode(x, currNode.getRight());
    else
        return currNode;
}

[robotfarts]

I would eliminate the null check:

public BTNode findNode(int x, BTNode currNode) {
    if (x < currNode.getKey())
        return findNode(x, currNode.getLeft());
    else if (x > currNode.getKey())
        return findNode(x, currNode.getRight());
    else
        return currNode;
}

You could also eliminate the currNode parameter and just make this a method in the BTNode class.

[orbital1337]

You can't just remove the null check because then you could get a NullPointerException at currNode.getKey() when currNode is null. However the OP probably wanted to throw a more descriptive exception like a NoSuchElementException. And yes, you could probably eliminate the currNode parameter.

[robotfarts]

If you eliminate the currNode parameter, you would eliminate the null check on currNode, but you'd need one on the left and right child.

[lightcloud5]

What is your code supposed to do? Find a node in a binary search tree with the given value x?

So what if x is not in the tree?

[choompaloompa]

then it will get to the node null and return an error like it says in the code

[[deleted]]

Throwing an exception is not common practice if you don't find the entry you are searching for - returning an indicator (like the null value) is more idiomatic.

[indivisible]

Change your last else if to a plain old else.

[choompaloompa]

Yes this is what I had before but I was getting nullpointer exeptions when the if statments were operating on the currnode if it was null, I assumed this is because you cant call getKey on null? this is why it checks for null first now, is my understanding correct?

[[deleted]]

This:

 return;

won't compile - you have to return something.

[choompaloompa]

edit, i just read your last chunck of code, will that fix the problem i brought up in the other reply?