Question

问题：将二叉搜索树应用于“person”对象而不是 int 值。人物对象由名称和重量组成。树将组织和搜索的person对象的值是一个人的“名称”（字符串）。

这是（叶子）树节点：

 public class Node {
   private Node leftChild;
   private Node rightChild;
   private Node parent;
   private Person person;
   private int height;

   public Node(Node p, Node l, Node r, Person paul, int h) {
     leftChild = l;
     rightChild = r;
     person = paul;
     height = h;
     parent = p;
   }

   public Node(Person paul, Node p) {
     this(p, null, null, paul, 0);
   }

   public Node (Person paul) {
     this(null, null, null, paul, 0);
   }

   public boolean isBalanced() {
     if(leftChild == null && rightChild == null)
       return true;
     else if (leftChild == null)
       return rightChild.getHeight() == 0;
     else if (rightChild == null)
       return leftChild.getHeight() == 0;

      return Math.abs(leftChild.getHeight()- rightChild.getHeight()) < 2;
   }

   public void setParent(Person parent) {
     parent = new Person( parent.getName(), parent.getWeight());
   }

   public Node getParent() {
     return parent;
   }

   public void setLeftChild(Node leftChild) {
     this.leftChild = leftChild;
   }

   public Node getLeftChild() {
     return leftChild;
   }

   public void setRightChild(Node rightChild) {
     this.rightChild = rightChild;
   }

   public Node getRightChild() {
     return rightChild;
   }

    public void setPerson(Person person) {
      Person = new Person(person.getName(), person.getWeight());
    }

    public Person getPerson() {
      return new Person(person.getName(), person.getWeight());
    }

 public void setHeight() {
  if(leftChild == null && rightChild == null)
   height = 0;
  else if (leftChild == null)
   height = rightChild.getHeight() + 1;
  else if (rightChild == null)
   height = leftChild.getHeight() + 1;
  else
   height= leftChild.getHeight() >= rightChild.getHeight()?
     leftChild.getHeight() + 1: rightChild.getHeight() +1;
 }

 public int getHeight() {
  setHeight();
  return height;
 }
}

这是BinarySearchTree：

public class BinarySearchTree {
 private Node root;

 public BinarySearchTree(Node root) {
  this.root = root;
 }
 public BinarySearchTree() {
  this.root = (Node)null;
 }

 void setRoot(Node r) {
  root = r;
 }

 Node getRoot() {
  return root;
 }

 public Node findParent(Person person, Node node) {
  if (node.getPerson() == person)//Error in code here
   return (Node)null; // root itself
  else if(node.getLeftChild() != null && node.getLeftChild().getPerson() == person)
   return node;
  else if(node.getRightChild() != null && node.getRightChild().getPerson() == person)
   return node;
  else if (node.getPerson().getName().compareTo(" a ")  > 0 ){
   return findParent(person, node.getLeftChild());}
  else
   return findParent(person, node.getRightChild());
 }

 public Node insertNode(Person person) {
  return insertNode(person, null, null);
 }

 public Node insertNode(Person person, Node node, Node parent) {
  if(node == (Node)null)
   node = new Node(parent, null, null, person, 0);
  else if (person.getName().compareTo(node.getPerson().getName()) < 0)
   node.setLeftChild(insertNode(person,node.getLeftChild(), node));
  else if (person.getName().compareTo(node.getPerson().getName()) > 0)
   node.setRightChild(insertNode(person, node.getRightChild(), node));
  node.setHeight();
  return node;
 }

 /* in-order traversal for showing inside of tree */
 public void traverseInOrder(Node node) {
  if(node.getLeftChild() != (Node)null)
   traverseInOrder(node.getLeftChild());
  System.out.print("Value: " + '\n' + node.getPerson().toString() + ", Height: " +
    node.getHeight() + ", Parent: ");

  Node n = findParent(node.getPerson(), getRoot());
  if(n == (Node)null)
   System.out.println("root");
  else
   System.out.println(n.getPerson().toString() + "");

  if(node.getRightChild()!= (Node)null)
   traverseInOrder(node.getRightChild());
 }
}




     public Node(Person paul, Node p) {
      this(p, null, null, paul, 0);

     }
     public Node (Person paul) {
      this(null, null, null, paul, 0);
     }

     public boolean isBalanced() {
      if(leftChild == null && rightChild == null)
       return true;
      else if (leftChild == null)
       return rightChild.getHeight() == 0;
      else if (rightChild == null)
       return leftChild.getHeight() == 0;

      return Math.abs(leftChild.getHeight()- rightChild.getHeight()) < 2;
     }

     public void setParent(Person parent) {
      parent = new Person( parent.getName(), parent.getWeight());

     //or is the parent supposed to be a null pointer ????
     }

     public Node getParent() {
      return parent;
     }

     public void setLeftChild(Node leftChild) {
          this.leftChild = leftChild;
         }

         public Node getLeftChild() {
          return leftChild;
         }

         public void setRightChild(Node rightChild) {
          this.rightChild = rightChild;
         }

         public Node getRightChild() {
          return rightChild;
         }

     public void setPerson(Person Person) {
     Person = new Person(person.getName(), person.getWeight());
     }

     public Person getPerson() {
         return new Person(person.getName(), person.getWeight());

     }

     public void setHeight() {
      if(leftChild == null && rightChild == null)
       height = 0;
      else if (leftChild == null)
       height = rightChild.getHeight() + 1;
      else if (rightChild == null)
       height = leftChild.getHeight() + 1;
      else
       height= leftChild.getHeight() >= rightChild.getHeight()?
         leftChild.getHeight() + 1: rightChild.getHeight() +1;
     }

     public int getHeight() {
      setHeight();
      return height;
     }

}

Answer 1

将String.compareTo()与人名作为参数。

Answer 2

问题在于

if (node.getPerson() == person)

测试这些对象在内存中是否具有相同的地址，而不是它们具有相同的“逻辑”值。您应该替换为：

if (node.getPerson().getName().equals(person.getName()))

BST数据结构 - 类项目 - 访问嵌套类

2 个答案: