java - 决策树执行

Question

我正在使用二元决策树开发一项调查。用户会收到一个问题，他们回答"is"或“否”，答案决定接下来要问什么问题。

我的问题是代码只能在左侧工作，这意味着用户只能回答“否”，代码才能正常工作。如果用户给出的答案是"is"，则会抛出 NullPointerException。难道是我做错了什么

    import java.util.Scanner;

    public class MusicGenreExpert {
    private static LinkedBinaryTree<String> tree;
    static Scanner scan = new Scanner(System.in);

    public MusicGenreExpert() throws ElementNotFoundException
    {
        String e1 = "Are you eco-friendly?";
        String e2 = "Do you consider yourself to be outgoing?";
        String e3 = "Do you consider yourself to be an introvert?";
        String e4 = "On an average day, do you think that you approach and solve" +
                     "problems differently than others?";
        String e5 = "Are you often uneasy (nervous)?";
        String e6 = "If the person in front of you dropped money without knowing," +
        "would you pick it up and keep it?";
        String e7 = "Is gentle a good word that describes you?";
        String e8 = "Do you often worry about what other people may think?";
        String e9 = "Do you think highly of yourself, with little to no care of" +
                      "what others may think?";
        String e10 = "Are you happy with the accomplishments you've had in life" +
                      "thus far?";
        String e11 = "Would you consider yourself to be old fashion?";
        String e12 = "Do you put your all in everything that you do?";
        String e13 = "Is your idea of an epic night going to a rave with thousands" +
                     "of people?";
        String e14 = "Do you believe that Bob Marley is one of the greatest artist" + 
                     "of all time?";
        String e15 = "Would you call yourself an extravert?";
        String e16 = "Do you know remember when you fell in love with" +
                     "Hip Hop?";
        String e17 = "You are the music genre classical. You are a creative introvert"+
        "with a good sense of self esteem. You are at ease with yourself as well as"+
        "the world around you.";
        String e18 = "The indie and punk music genre best fits you. You are an intellectual" +
        "and creative introvert. You prefer not to work as hard as others and are less gentle." +
        "Your self esteem is a constant battle.";
        String e19 = "Rock/Heavy Metal best describes you. You are a creative and gentle" +
        "introvert with a constant battle with self esteem. However you are no stranger to" +
        "hard work.";
        String e20 = "You are a high self esteem, honest and convetional" +
        "introvert. The Pop culture is who you are inside. You may lack in creativity," +
        "but you make up for in hard work.";
        String e21 = " You remember the exact day you fell in love with Hip Hop." +
        "\n With your high self esteem and outgoing personality, you are sure to be" +
        "\n the life of the party. However, you could be more eco friendly and less" +
        " aggressive.";
        String e22 = "You are jazz, blues, and soul. You are a lot more extraverted" +
        "than others. Your at ease personality, high self esteem, and creativity" +
        "makes you the person that eveyone wants to know.";
        String e23 = "The electronic/dance genre describes exactly who you are." +
        "Assertive, outgoing, creative, yet aggressive gives everyone that comes" +
        "in contact with you something to love. You aren't as kind as others, but" +
        "hey, they love you anyway.";
        String e24 = "Music is suppose to tell a story, and country music does" +
        "just that. You are hardworking, conventional and outgoing. Oh and contrary" +
        "to what others may believe your emotional stability is just fine.";
        String e25 = "There is no other genre better to describe you than Reggae." +
        "You are a creative outgoing individual, with high self esteem. Where you" +
        "lack in hardwork you make up for in your kind and generous ways.";

        LinkedBinaryTree<String> n2, n3, n4, n5, n6, n7, n8, n9, n10, n11, n12, n13, n14,
        n15, n16, n17, n18, n19, n20, n21, n22, n23, n24, n25, n26, n27, n28, n29, n30,
        n31, n32, n33, n34, n35, n36, n37, n38, n39, n40, n41, n42, n43, n44, n45, n46,
        n47, n48, n49, n50, n51, n52, n53;

        n50 = new LinkedBinaryTree<String> (e16);
        n51 = new LinkedBinaryTree<String> (e17);
        n33 = new LinkedBinaryTree<String> (e8, n50, n51);

        n52 = new LinkedBinaryTree<String> (e16);
        n53 = new LinkedBinaryTree<String> (e18);
        n49 = new LinkedBinaryTree<String> (e8, n52, n53);

        n32 = new LinkedBinaryTree<String> (e21);
        n16 = new LinkedBinaryTree<String> (e15, n32, n33);

        n34 = new LinkedBinaryTree<String> (e17);
        n35 = new LinkedBinaryTree<String> (e18);
        n18 = new LinkedBinaryTree<String> (e7, n34, n35);

        n36 = new LinkedBinaryTree<String> (e18);
        n37 = new LinkedBinaryTree<String> (e19);
        n19 = new LinkedBinaryTree<String> (e12, n36, n37);

        n38 = new LinkedBinaryTree<String> (e16);
        n39 = new LinkedBinaryTree<String> (e20);
        n23 = new LinkedBinaryTree<String> (e16, n38, n39);

        n40 = new LinkedBinaryTree<String> (e21);
        n41 = new LinkedBinaryTree<String> (e23);
        n25 = new LinkedBinaryTree<String> (e8, n40, n41);

        n42 = new LinkedBinaryTree<String> (e22);
        n43 = new LinkedBinaryTree<String> (e20);
        n26 = new LinkedBinaryTree<String> (e16, n42, n43);

        n44 = new LinkedBinaryTree<String> (e23);
        n45 = new LinkedBinaryTree<String> (e19);
        n27 = new LinkedBinaryTree<String> (e6, n44, n45);

        n46 = new LinkedBinaryTree<String> (e17);
        n47 = new LinkedBinaryTree<String> (e24);
        n28 = new LinkedBinaryTree<String> (e14, n46, n47);

        n48 = new LinkedBinaryTree<String> (e17);
        n30 = new LinkedBinaryTree<String> (e7, n48, n49);

        n17 = new LinkedBinaryTree<String> (e18);
        n8 = new LinkedBinaryTree<String> (e7, n16, n17);

        n9 = new LinkedBinaryTree<String> (e9, n18, n19);

        n20 = new LinkedBinaryTree<String> (e17);
        n21 = new LinkedBinaryTree<String> (e18);
        n10 = new LinkedBinaryTree<String> (e7, n20, n21);

        n22 = new LinkedBinaryTree<String> (e19);
        n11 = new LinkedBinaryTree<String> (e5, n22, n23);

        n24 = new LinkedBinaryTree<String> (e19);
        n12 = new LinkedBinaryTree<String> (e10, n24, n25);

        n13 = new LinkedBinaryTree<String> (e11, n26, n27);

        n29 = new LinkedBinaryTree<String> (e19);
        n14 = new LinkedBinaryTree<String> (e12, n28, n29);

        n31 = new LinkedBinaryTree<String> (e22);
        n15 = new LinkedBinaryTree<String> (e13, n30, n31);

        n4 = new LinkedBinaryTree<String> (e6, n8, n9);
        n5 = new LinkedBinaryTree<String> (e9, n10, n11);

        n6 = new LinkedBinaryTree<String> (e4, n12, n13);
        n7 = new LinkedBinaryTree<String> (e3, n14, n15);

        n2 = new LinkedBinaryTree<String> (e2, n4, n5);
        n3 = new LinkedBinaryTree<String> (e3, n6, n7);

        tree = new LinkedBinaryTree<String> (e1, n2, n3);
    }

    public static void diagnose(){
        BinaryTree<String> current = tree;

        System.out.println("                        What music genre are you?       ");
        while(current.size() > 1)
        {
            System.out.println (current.getRootElement());
            if (scan.nextLine().equalsIgnoreCase("N"))
                current = current.getLeft();
            else
                current =  current.getRight();  
        }

        System.out.println (current.getRootElement());          
    }      

}

这是实现 BinaryTree 的 LinkedBinaryTree:

import java.util.Iterator;

public class LinkedBinaryTree<T> implements BinaryTree<T>
{
   protected BTNode<T> root;

   public LinkedBinaryTree(){
    root = null; 
 }

  public LinkedBinaryTree (T element)
  {
     root = new BTNode<T>(element);
  }

  public LinkedBinaryTree (T element, LinkedBinaryTree<T> left,
          LinkedBinaryTree<T> right)
       {
          root = new BTNode<T>(element);
          root.setLeft(left.root);
          root.setRight(right.root);
       }

   public T getRootElement()
   {
      if (root == null)
      throw new EmptyCollectionException ("Get root operation "
        + "failed. The tree is empty.");

      return root.getElement();
   }

   public LinkedBinaryTree<T> getLeft()
   {
      if (root == null)
      throw new EmptyCollectionException ("Get left operation "
        + "failed. The tree is empty.");

      LinkedBinaryTree<T> result = new LinkedBinaryTree<T>();
      result.root = root.getLeft();

       return result;
    }

    public T find (T target)
   {
      BTNode<T> node = null;

      if (root != null)
      node = root.find(target);

      if (node == null)
      throw new ElementNotFoundException("Find operation failed. "
        + "No such element in tree.");

      return node.getElement();
    }

    public int size()
   {
      int result = 0;

      if (root != null)
     result = root.count();

      return result;
   }

   public Iterator<T> iterator() 
   {
    return new ArrayIterator<T>();
    }

  public int height()
   {
     int result = 0;

     if (root == null)
      return result;

     if (root != null)
     result = root.height();  
     return result; 
  }

public BinaryTree<T> getRight() {
// TODO Auto-generated method stub
return null;
}

public boolean contains(T target) {
// TODO Auto-generated method stub
  return false;
}

 public boolean isEmpty() {
// TODO Auto-generated method stub
  return false;
  }

}

Answer 1

现在您已经提供了 LinkedBinaryTree 的源代码，问题就清楚了。唯一一次调用 getRight() 是当您对其中一个决策回答 Y 或"is"时，该决策将执行这段代码:

public BinaryTree<T> getRight() {
  // TODO Auto-generated method stub
  return null;
}

由于它返回 null，因此下次尝试取消引用 current 时(这发生在代码 while(current.size() > 1) 中)，它会导致 NullPointerException。

getRight() 方法需要类似于 getLeft() 实现。