Question

我正在查询树，查找带有数据字符串“directory”的节点，然后在列表中报告其子项的数据。

在我的方法开始时，我已经将我的列表初始化了。列表listofdirectories = new ArrayList＆lt;＆gt;（）;

在调试时，java会递归遍历树并将字符串添加到列表中，但是一旦它再次调用自身，listofdirectories变量在递归期间将再次变为空。但是，最后，它会列出根目录及其目录子目录。

我的方法没有参数。在pastebin上发布了代码，有很多System.out供我调试。

我的方法代码+控制台+示例树发布在这里： https://pastebin.com/9nXvcCNS

public List<String> getDirectories() // WORKS b/c didnt add first directory
        {
                List<String> listofdirectories = new ArrayList<>();
                System.out.println("AT THE START OF THE METHOD " +listofdirectories);
               // int sizeofchildrennodes = this.children.size();

                if(this.data.equals("directory"))
                {
                    System.out.println("FFFFFFFFFFFFFFF");
                }

                if(this.data.equals(("directory")))
                {
                    listofdirectories.add(this.children.get(0).children.get(0).data);

                for (int i = 0; i < this.children.size(); i++) { // Checks the directories in the CURRENT directory

                    String nodename = this.children.get(i).data;              
                    FileSystemNode thenode = this.children.get(i);

                    System.out.println("Current node is: " + nodename);
                    System.out.println("Current nodedata is: " + thenode.children.get(0).data);

                  if(nodename.equals("directory2.0"))
                {
                    System.out.println("I found direct2.0");
                    System.out.println("name of direct2.0 is " + thenode.children.get(0).children.get(0).data);
                    System.out.println("0" + listofdirectories);
                    listofdirectories.add(thenode.children.get(0).children.get(0).data);
                    System.out.println("0).5" +listofdirectories);
                }

                    System.out.println("1" +listofdirectories);

                if(nodename.equals("directory"))
                {
                    System.out.println("@@" + listofdirectories);
                    listofdirectories.add(thenode.children.get(0).children.get(0).data);
                    if(thenode.children.size()> 0)
                    {
                        thenode.getDirectories();
                    }

                }

                }
//                if(this.children.get(i).children.contains(i))

            }
                System.out.println("!!!beforeoutput" +listofdirectories);
                return listofdirectories;
        }


//////////////////////////////////
/// This is console's log
////////////////////////////////////
AT THE START OF THE METHOD []
FFFFFFFFFFFFFFF
Current node is: name
Current nodedata is: Movies and Pictures
1[Movies and Pictures]
Current node is: directory
Current nodedata is: name
1[Movies and Pictures]
@@[Movies and Pictures]
AT THE START OF THE METHOD []
FFFFFFFFFFFFFFF
Current node is: name
Current nodedata is: Movies
1[Movies]
Current node is: file
Current nodedata is: name
1[Movies]
Current node is: file
Current nodedata is: name
1[Movies]
Current node is: file
Current nodedata is: name
1[Movies]
!!!beforeoutput[Movies]
Current node is: directory
Current nodedata is: name
1[Movies and Pictures, Movies]
@@[Movies and Pictures, Movies]
AT THE START OF THE METHOD []
FFFFFFFFFFFFFFF
Current node is: name
Current nodedata is: Pictures
1[Pictures]
Current node is: file
Current nodedata is: name
1[Pictures]
Current node is: file
Current nodedata is: name
1[Pictures]
Current node is: file
Current nodedata is: name
1[Pictures]
Current node is: directory2.0
Current nodedata is: name
I found direct2.0
name of direct2.0 is Pictures2.0
0[Pictures]
0).5[Pictures, Pictures2.0]
1[Pictures, Pictures2.0]
!!!beforeoutput[Pictures, Pictures2.0]
!!!beforeoutput[Movies and Pictures, Movies, Pictures]
2 [Movies and Pictures, Movies, Pictures] // THIS IS THE OUTPUT
//I WANT THIS TO INCLUDE PICTURES2.0

///////////////////////
//// Example of a tree
//////////////////////

|-- directory
    |-- name
        |-- Movies and Pictures
    |-- directory
        |-- name
            |-- Movies
        |-- file
            |-- name
                |-- Harry Potter.mov
            |-- date
                |-- 20130602
            |-- size
                |-- 1400123
        |-- file
            |-- name
                |-- Dodgeball.avi
            |-- date
                |-- 11111111
            |-- size
                |-- 5000000
        |-- file
            |-- name
                |-- The Dark Knight.mpeg
            |-- date
                |-- 55555555
            |-- size
                |-- 7657657
    |-- directory
        |-- name
            |-- Pictures
        |-- file
            |-- name
                |-- Mountain.png
            |-- date
                |-- 20000000
            |-- size
                |-- 190000
        |-- file
            |-- name
                |-- Flower.jpg
            |-- date
                |-- 140000
            |-- size
                |-- 7657657
        |-- file
            |-- name
                |-- Wildlife.jpeg
            |-- date
                |-- 90000000
            |-- size
                |-- 250000
        |-- directory2.0
            |-- name
                |-- Pictures2.0

解决

示例树查询的解决方案如下：https://pastebin.com/FkBTatP6

Answer 1

List<String> listofdirectories = new ArrayList<>();

在全球范围内宣布。

Answer 2

您没有对返回值执行任何操作。看这里：

thenode.getDirectories();

好的，您已经调用了该方法，但是如何处理其返回值集？没有。你没有存储它。那么，如果你没有存储它，为什么你希望你的代码返回正确的集合呢？我认为这应该让你更接近问题的解决方案：

listofdirectories.addAll(thenode.getDirectories());

递归java初始化列表为空

解决

2 个答案: