打印树并跳过某些值

Question

这里我有以下代码和打印树结构的输入。我的问题是如何才能使具有“不可用”值的节点和叶子被跳过打印。

namespace Tree{public class TreeNode<T>

{
    private T value;

    private bool hasParent;

    private List<TreeNode<T>> children;


    public TreeNode(T value)
    {
        if (value == null)
        {
            throw new ArgumentNullException("Cannot insert null value");
        }

        this.value = value;
        this.children = new List<TreeNode<T>>();
    }
    public T Value
    {
        get
        {
            return this.value;
        }
        set
        {
            this.value = value;
        }
    }

    public int ChildrenCount
    {
        get
        {
            return this.children.Count;
        }
    }
    public void AddChild(TreeNode<T> child)
    {
        if (child == null)
        {
            throw new ArgumentNullException("Cannot insert null value");
        }
        if (child.hasParent)
        {
            throw new ArgumentException("The node already has a parent");
        }

        child.hasParent = true;
        this.children.Add(child);
    }
    public TreeNode<T> GetChild(int index)
    {
        return this.children[index];
    }
}

public class Tree<T>
{

    private TreeNode<T> root;
    public Tree(T value)
    {
        if (value == null)
        {
            throw new ArgumentNullException("Cannot insert null value");
        }

        this.root = new TreeNode<T>(value);
    }

    public Tree(T value, params Tree<T>[] children) : this(value)
    {
        foreach (Tree<T> child in children)
        {
            this.root.AddChild(child.root);
        }
    }

    public TreeNode<T> Root
    {
        get
        {
            return this.root;
        }
    }

    private void PrintDFS(TreeNode<T> root, string spaces)
    {
        if (this.root == null)
        {
            return;
        }
        Console.WriteLine(spaces + root.Value);

        TreeNode<T> child = null;
        for (int i = 0; i < root.ChildrenCount; i++)
        {
            child = root.GetChild(i);
            PrintDFS(child, spaces + "   ");
        }
    }
    public void TraverseDFS()
    {
        this.PrintDFS(this.root, string.Empty);
    }
}

public static class TreeExample
{
    static void Main()
    {

        Tree<string> tree =
        new Tree<string>("John",
            new Tree<string>("Jasmine",
                new Tree<string>("Jay"),
                new Tree<string>("Unavailable")),
             new Tree<string>("Unavailable",
                new Tree<string>("Jack"),
                new Tree<string>("Jeremy")),
            new Tree<string>("Johanna")

            );
    tree.TraverseDFS();


    }
}}

现在它打印 :(John, (Jasmine, (Jay), (Unavailable)), (Unavailable, (Jack, (Jeremy))), (Johanna))

我需要它来打印 :(John, (Jasmine, (Jay)), (Johanna))

所以基本上跳过每个值为“Unavailable”的叶子和每个值为“Unavailable”的节点以及该节点的所有子节点

谢谢！

Answer 1

这应该可行：

private void PrintDFS(TreeNode<T> root, string spaces)
{
    if (this.root == null
        || "Unavailable" == root.Value.ToString())
    {
        return;
    }
    ...

Answer 2

接受的答案是对问题的字面正确答案，但它在逻辑上将关于如何处理树的逻辑融入到树本身中。树是一种集合或数据结构，您通常不会看到能够打印自身的列表或字典。相反，该集合提供了正确的方法来获取或更改其内容，以便您可以随心所欲。

在您的情况下，您可以执行以下操作：

public enum TreeVisitorResult {
    SkipNode,
    Continue
}

// the following two methods inside Tree<T>:

public void VisitNodes(Func<TreeNode<T>, int, TreeVisitorResult> visitor) {
    VisitNodes(0, this.root, visitor);
}

private void VisitNodes(int depth, TreeNode<T> node,
    Func<TreeNode<T>, int, TreeVisitorResult> visitor) {

    if (node == null) {
        return;
    }

    var shouldSkip = visitor(node, depth);
    if (shouldSkip == TreeVisitorResult.SkipNode) {
        return;
    }

    TreeNode<T> child = null;
    for (int i = 0; i < node.ChildrenCount; i++) {
        child = node.GetChild(i);
        VisitNodes(depth + 1, child, visitor);
    }
}

如果你有这个方法，你可以在 Tree 类之外编写 Print 方法，如下：

tree.VisitNodes((treeNode, depth) => { // <- this lambda will be called for every node
    if (treeNode.Value == "Unavailable") { // <- no need to ToString or cast here, since
                                           // we know that T is string here
        return TreeVisitorResult.SkipNode;
    } else {
        var spaces = new string(' ', depth * 3);
        Console.WriteLine(spaces + treeNode.Value);     
    }
});