Question

我想实现以下代码。要求如下

完成构造函数并测试您在方法main中执行的操作。
在方法main中，构建一个包含几个节点的树，然后为每个名称调用findChild。他们会失败。完成方法findChild直到测试通过。

我很困惑的一件事是f1的FileNode为null，并且null值不能放在hashmap中（将显示空指针错误）。我该怎么做才能防止这个问题。

这是我做的代码

package part2;

import java.util.Map;

import javax.swing.JFileChooser;

import java.io.File;
import java.util.Collection;


/**
 * The root of a tree representing a directory structure.
 */
public class FileNode {

    /** The name of the file or directory this node represents. */
    private String name;
    /** Whether this node represents a file or a directory. */
    private FileType type;
    /** This node's parent. */
    private FileNode parent;
    /**
     * This node's children, mapped from the file names to the nodes. If type is
     * FileType.FILE, this is null.
     */
    private Map<String, FileNode> children;

    /**
     * A node in this tree.
     *
     * @param name
     *            the file
     * @param parent
     *            the parent node.
     * @param type
     *            file or directory
     * @see buildFileTree
     */
    public FileNode(String name, FileNode parent, FileType type) {
        this.name = name;
        this.parent = parent;
        this.type = type;



        // TODO: complete this method.
    }

    /**
     * Find and return a child node named name in this directory tree, or null
     * if there is no such child node.
     *
     * @param name
     *            the file name to search for
     * @return the node named name
     */
    public FileNode findChild(String name) {
        // TODO: complete this method.
            if (children.containsKey(name)) {
                return children.get(name);
            } 
            else{
                return null;
            }

    }

    /**
     * Return the name of the file or directory represented by this node.
     *
     * @return name of this Node
     */
    public String getName() {
        return this.name;
    }

    /**
     * Set the name of the current node
     *
     * @param name
     *            of the file/directory
     */
    public void setName(String name) {
        this.name = name;
    }

    /**
     * Return the child nodes of this node.
     *
     * @return the child nodes directly underneath this node.
     */
    public Collection<FileNode> getChildren() {
        return this.children.values();
    }

    /**
     * Return this node's parent.
     * 
     * @return the parent
     */
    public FileNode getParent() {
        return parent;
    }

    /**
     * Set this node's parent to p.
     * 
     * @param p
     *            the parent to set
     */
    public void setParent(FileNode p) {
        this.parent = p;
    }

    /**
     * Add childNode, representing a file or directory named name, as a child of
     * this node.
     * 
     * @param name
     *            the name of the file or directory
     * @param childNode
     *            the node to add as a child
     */
    public void addChild(String name, FileNode childNode) {
        this.children.put(name, childNode);
    }

    /**
     * Return whether this node represents a directory.
     * 
     * @return whether this node represents a directory.
     */
    public boolean isDirectory() {
        return this.type == FileType.DIRECTORY;
    }

    /**
     * This method is for code that tests this class.
     * 
     * @param args
     *            the command line args.
     */
    public static void main(String[] args) {
        JFileChooser fileChooser = new JFileChooser();
        fileChooser.setFileSelectionMode(JFileChooser.FILES_ONLY);
        int returnVal = fileChooser.showOpenDialog(null);
        File file;

        if (returnVal == JFileChooser.APPROVE_OPTION) {
            file = fileChooser.getSelectedFile();
            System.out.println(file);
        }
        System.out.println("Testing FileNode");
        FileNode f1 = new FileNode("top", null, FileType.FILE);
        FileNode f2 = new FileNode("top", f1, FileType.FILE);

        f2 = file;
        f1.addChild("c1", f2);

        System.out.println(f2.findChild("c1"));
        if (!f1.getName().equals("top")) {
            System.out.println("Error: " + f1.getName() + " should be " + "top");
        }


    }

}

以下是错误消息：

Exception in thread "main" java.lang.NullPointerException
    at part2.FileNode.addChild(FileNode.java:126)
    at part2.FileNode.main(FileNode.java:152)

Answer 1

f1的FileNode为null，并且无法将空值放入hashmap

我认为null可以放在HashMap中，至少作为值。

错误意味着children == null，因为你从未初始化它！所以解决这个问题......

public void addChild(String name, FileNode childNode) {
    if (this.children == null) {
        this.children = new HashMap<String, FileNode>();
    }
    this.children.put(name, childNode);
}

但是，当你找到一个孩子时，你只是在检查直系孩子。您需要迭代文件树，直到找到所需内容。

public FileNode findChild(String name) {
    // Base case - no children
    if (children == null) return null;

    // Get the child, if exists
    FileNode immediateChild = children.get(name);
    if (immediateChild != null) {
        return children.get(name);
    } 
    else { // Iterate all children until found
        FileNode tmp = null;
        for (FileNode f : getChildren()) {
            if (tmp != null) break;
            tmp = f.getChild(name);
        }
        return tmp;
    }
}

使用findChild方法构建文件树

1 个答案: