使用递归查找从根节点到树中指定节点的路径

Question

背景：这是我第一次使用树，我的任务是使用文件 data.txt 使用皇室创建一棵树

数据.txt：

King George VI
King George VI > Princess Margaret
Princess Margaret > David, Viscount Linley
Princess Margaret > Lady Sarah
Lady Sarah > Samuel Chatto
Lady Sarah > Arthur Chatto
David, Viscount Linley > Charles Armstrong-Jones
David, Viscount Linley > Margarita Armstrong-Jones
King George VI > Queen Elizabeth II
Queen Elizabeth II > Charles, Prince of Wales
Charles, Prince of Wales > Prince William of Wales
Prince William of Wales > Prince George of Cambridge
Charles, Prince of Wales > Prince Harry of Wales
Queen Elizabeth II > Anne, Princess Royal
Anne, Princess Royal > Peter Phillips
Peter Phillips > Savannah Phillips
Peter Phillips > Isla Phillips
Anne, Princess Royal > Zara Tindall
Queen Elizabeth II > Andrew, Duke of York
Andrew, Duke of York > Princess Beatrice of York
Andrew, Duke of York > Princess Eugenie of York
Queen Elizabeth II > Edward, Earl of Wessex
Edward, Earl of Wessex > Lady Louise Windsor
Edward, Earl of Wessex > James, Viscount Severn

背景（Cotd.） ：然后我搜索约克公主比阿特丽斯，并找到她各自的节点。

    public static void main(String[] args)
    {
        //Define a variable to store the root node
        TNode<String> root = null;

        //TODO: SETUP TREE DATA
        //1. Use Scanner to read the data.txt file
        //2. The first line in data.txt is the root node
        //3. For each line in data.txt in the format A > B
        //      - *find* the A node
        //      - add B as a child of A
        try
        {
            Scanner s = new Scanner(new File("data.txt"));
            while(s.hasNextLine())
            {
                String[] split = null;
                if(!s.nextLine().contains(">"))
                {
                    root = new TNode<String>(s.nextLine());
                }else{
                     split = s.nextLine().split(" > ");
                }
                find(root,split[0]).setParent(new TNode(split[3]));
            }
            s.close();
        }catch(Exception e){
            e.printStackTrace();
        }
        //TODO: test printPath method
        TNode<String> child = find(root, "Princess Beatrice of York");
        String path = getPath( child );
        System.out.println(path);

    }

查找方法：

    /**
     *  @return node if its data matches name, or return a child node with data that matches name
     */
    public static TNode<String> find(TNode<String> node, String name)
    {
        //use recursion to check this node and all of its children to see if their data matches the specified name
        if(node.getData().equals(name))
        {
            return node;
        }
        for(int i = 0; i < node.getChildren().size(); i++)
        {
            return find(node.getChildren().get(i),name);
        }
        return null;
    }

目标：返回一个包含从根节点到指定节点的路径的字符串，以' -> '分隔，从子节点开始

我的尝试：

    public static String getPath(TNode<String> node)
    {
        //use recursion to concatenate the getPath of this node's parent with this node's data
        if(!node.getParent().equals(null))
            return getPath(node.getParent()) + " -> " + node.getData(); 
        return getPath(node.getParent());
    }

问题：

1.

这甚至可能首先

2.

如果没有可以连接的字符串，我该如何连接字符串。

3.

有哪些方法可以帮助我找出递归问题的逻辑

Answer 1

这是绝对可能的。 单看它，就可以追溯到从Princess Beatrice of York到King George VI 。

要创建家庭行字符串（假设这是您想要的），您可以将每个人存储在一个TNode ，并从根开始用>将名字连接起来。

至于学习考虑递归算法，从更简单的事情开始，比如字符串的长度或计算数字的阶乘。