CODEWITHSUNDEEP
javatreereturn
juancsr
juancsr

Reputation: 329

Java return statement priority

I have this code:

private String buscar(Nodo nodo, Nodo raiz) {
    System.out.println(nivel);
    this.nivel++;

    if (raiz != null) {
        if (nodo.getValor() == raiz.getValor()) {
            System.out.println("es igual");
            return "El número " + nodo.getValor() + " está en el nivel: " + this.nivel;

        } else if (nodo.getValor() < raiz.getValor()) {
            buscar(nodo, raiz.getIzq());

        } else {
            System.out.println("es mayor");
            buscar(nodo, raiz.getDer());

        }
    } else {
        return "El número " + nodo.getValor() + " no forma parte del árbol";
    }

    return null;

}

public String buscar(int valor) {
    this.nivel = 0;
    Nodo nodo = new Nodo(valor);
    return this.buscar(nodo, this.raiz);
}

and is always execute the last return sentence (return null), even when the if sentence becomes true. Can anybody tell me why this is happening?

Upvotes: 0

Views: 157

Answers (1)

Elliott Frisch
Elliott Frisch

Reputation: 201467

You need to return the result of your recursive calls, you can also eliminate two of the else conditions (since you return). And, I would prefer String.format over concatenation. Like,

private String buscar(Nodo nodo, Nodo raiz) {
    System.out.println(nivel);
    this.nivel++;

    if (raiz != null) {
        if (nodo.getValor() == raiz.getValor()) {
            System.out.println("es igual");
            return String.format("El número %d está en el nivel: %d", 
                        nodo.getValor(), this.nivel);
        } else if (nodo.getValor() < raiz.getValor()) {
            return buscar(nodo, raiz.getIzq()); // return the recursive result
        }
        System.out.println("es mayor");
        return buscar(nodo, raiz.getDer()); // return the recursive result
    }
    return String.format("El número %d no forma parte del árbol", nodo.getValor());
}

Upvotes: 1

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