Java：toString方法打印哈希码而不是inOrder遍历

发布于 2025-01-10 12:40:00 字数 1264 浏览 0 评论 0原文

我正在尝试打印 BinarySearchTree(Generic) 的 toString，其主体中包含以下内容：

@Override
    public String toString() {
    root.inOrderTraversal(root);
    return  "inOrderTraversal has finished";
    }

这是我的 inOrder 遍历，它位于 BinaryNode 类 (Generic) 内部，由 BinarySearchTree 使用：


public void inOrderTraversal(BinaryNode<T> node)
    {
        if(node != null)
        {
            if(node.left != null)
            {
                inOrderTraversal(node.left);
            }
            System.out.println(node.nodeValue);
            if(node.right != null)
            {
                inOrderTraversal(node.right);
            }
        }
    }

在我使用Student 作为其类型并打印了 toString，它显示输出为 Student@7a81197d、Student@5ca881b5，学生@24d46ca6。可能是什么问题？


        Student s1 = new Student("Jasim", 84812); //Name and his/her ID
        Student s2 = new Student("Yousef", 845623);
        Student s3 = new Student("Zack", 432553);
        Student s4 = new Student("Zara", 54233);
        BinarySearchTree<Student> bst = new BinarySearchTree<Student>(s1); //construction
        bst.insert(s2);
        bst.insert(s3); 
        System.out.println(bst.toString()); //Error when printing this.

原文

I am trying to print the toString of BinarySearchTree(Generic), which contains in its body the following:

@Override
    public String toString() {
    root.inOrderTraversal(root);
    return  "inOrderTraversal has finished";
    }

and this is my inOrder traversal, which is inside the BinaryNode class (Generic), used by the BinarySearchTree:


public void inOrderTraversal(BinaryNode<T> node)
    {
        if(node != null)
        {
            if(node.left != null)
            {
                inOrderTraversal(node.left);
            }
            System.out.println(node.nodeValue);
            if(node.right != null)
            {
                inOrderTraversal(node.right);
            }
        }
    }

After I constructed the Generic BinarySerachTree using Student as its type and printed the toString, it is showing the output as Student@7a81197d, Student@5ca881b5, Student@24d46ca6.
What could be the problem?


        Student s1 = new Student("Jasim", 84812); //Name and his/her ID
        Student s2 = new Student("Yousef", 845623);
        Student s3 = new Student("Zack", 432553);
        Student s4 = new Student("Zara", 54233);
        BinarySearchTree<Student> bst = new BinarySearchTree<Student>(s1); //construction
        bst.insert(s2);
        bst.insert(s3); 
        System.out.println(bst.toString()); //Error when printing this.

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笙痞 2025-01-17 12:40:00

如果你的导师维持原来的要求，你可以尝试以下方法：
`

public class StudentModal extends Student{
    public StudentModal( String name, Integer id) {
        super(name, id);
    }

    @Override
    public String toString() {
        return "userName:" + this.getName() + "id:" +this.getId();
    }
}
public static void main(String[] args) {
    StudentModal student = new StudentModal("test", 1);
    System.out.println(student);
}

If your instructor maintains the original requirements, you can try the following：
`

public class StudentModal extends Student{
    public StudentModal( String name, Integer id) {
        super(name, id);
    }

    @Override
    public String toString() {
        return "userName:" + this.getName() + "id:" +this.getId();
    }
}
public static void main(String[] args) {
    StudentModal student = new StudentModal("test", 1);
    System.out.println(student);
}

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