总结所有节点

发布于 2024-07-08 07:08:56 字数 548 浏览 8 评论 0原文

这可能是一个简单的修复 - 但我试图将二叉搜索树上的所有节点（Node 类的 Size 属性）相加。到目前为止，在我的 BST 类中，我有以下内容，但它返回 0：

    private long sum(Node<T> thisNode)
    {
        if (thisNode.Left == null && thisNode.Right == null)
            return 0;
        if (node.Right == null)
            return sum(thisNode.Left);
        if (node.Left == null) 
            return sum(thisNode.Right);


        return sum(thisNode.Left) + sum(thisNode.Right);
    }

在我的 Node 类中，我有 Data，它将 Size 和 Name 存储在其给定属性中。我只是想总结整个大小。有什么建议或想法吗？

原文

This may be a simple fix - but I'm trying to sum together all the nodes (Size property from the Node class) on the binary search tree. Below in my BST class I have the following so far, but it returns 0:

    private long sum(Node<T> thisNode)
    {
        if (thisNode.Left == null && thisNode.Right == null)
            return 0;
        if (node.Right == null)
            return sum(thisNode.Left);
        if (node.Left == null) 
            return sum(thisNode.Right);


        return sum(thisNode.Left) + sum(thisNode.Right);
    }

Within my Node class I have Data which stores Size and Name in their given properties. I'm just trying to sum the entire size. Any suggestions or ideas?

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沫离伤花 2024-07-15 07:08:56

这是因为当您到达叶节点时，您将返回零。您应该返回存储在该叶节点中的大小。

此外，如果您的非叶节点也有大小，您也需要这样处理它们：

private long sum(Node<T> thisNode)
{
    if (thisNode.Left == null && thisNode.Right == null)
        return thisNode.Size;
    if (node.Right == null)
        return thisNode.Size + sum(thisNode.Left);
    if (node.Left == null) 
        return thisNode.Size + sum(thisNode.Right);
    return thisNode.Size + sum(thisNode.Left) + sum(thisNode.Right);
}

如果您的非叶节点没有大小，请使用：

private long sum(Node<T> thisNode)
{
    if (thisNode.Left == null && thisNode.Right == null)
        return thisNode.Size;
    if (node.Right == null)
        return sum(thisNode.Left);
    if (node.Left == null) 
        return sum(thisNode.Right);
    return sum(thisNode.Left) + sum(thisNode.Right);
}

第一个的更优雅的版本是：

private long sum(Node<T> thisNode)
{
    if (thisNode == null)
        return 0;
    return thisNode.Size + sum(thisNode.Left) + sum(thisNode.Right);
}

It's because you're returning zero when you reach a leaf node. You should be returning the size stored in that leaf node.

In addition, if your non-leaf nodes also have a size, you'll need to process them as well thus:

private long sum(Node<T> thisNode)
{
    if (thisNode.Left == null && thisNode.Right == null)
        return thisNode.Size;
    if (node.Right == null)
        return thisNode.Size + sum(thisNode.Left);
    if (node.Left == null) 
        return thisNode.Size + sum(thisNode.Right);
    return thisNode.Size + sum(thisNode.Left) + sum(thisNode.Right);
}

If your non-leaf nodes don't have size, use:

private long sum(Node<T> thisNode)
{
    if (thisNode.Left == null && thisNode.Right == null)
        return thisNode.Size;
    if (node.Right == null)
        return sum(thisNode.Left);
    if (node.Left == null) 
        return sum(thisNode.Right);
    return sum(thisNode.Left) + sum(thisNode.Right);
}

A more elegant version of the first one is:

private long sum(Node<T> thisNode)
{
    if (thisNode == null)
        return 0;
    return thisNode.Size + sum(thisNode.Left) + sum(thisNode.Right);
}

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咋地 2024-07-15 07:08:56

也许你的意思是

    if (thisNode.Left == null && thisNode.Right == null)
        return thisNode.Size;

？

Maybe you meant

    if (thisNode.Left == null && thisNode.Right == null)
        return thisNode.Size;

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凉薄对峙 2024-07-15 07:08:56

试试这个：

    private long sum(Node<T> thisNode)
    {
        if (thisNode == null)
            return 0;
        return thisNode.Size + sum(thisNode.Left) + sum(thisNode.Right);
    }

原始代码返回的唯一“值”是 0 - 这就是结果始终为 0 的原因。

Try this:

    private long sum(Node<T> thisNode)
    {
        if (thisNode == null)
            return 0;
        return thisNode.Size + sum(thisNode.Left) + sum(thisNode.Right);
    }

The only "value" that the original code ever returns is 0 - that's why the result is always 0.

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酒解孤独 2024-07-15 07:08:56

？

private long Sum(Node<T> thisNode)
{
  if( thisNode == null )
    return 0;

  return thisNode.Size + Sum(thisNode.Left) + Sum(thisNode.Right);
}

如果尺寸属性不在节点本身上，那怎么样呢

    public class Node<T>
    {
        public T Data;
        public Node<T> Left;
        public Node<T> Right;

        public static void ForEach(Node<T> root, Action<T> action)
        {
            action(root.Data);

            if (root.Left != null)
                ForEach(root.Left, action);
            if (root.Right != null)
                ForEach(root.Right, action);
        }
    }

    public interface IHasSize
    {
        long Size { get; }
    }

    public static long SumSize<T>(Node<T> root) where T : IHasSize
    {
        long sum = 0;
        Node<T>.ForEach(root, delegate(T item)
        {
            sum += item.Size;
        });
        return sum;
    }

How about

private long Sum(Node<T> thisNode)
{
  if( thisNode == null )
    return 0;

  return thisNode.Size + Sum(thisNode.Left) + Sum(thisNode.Right);
}

If the size property isn't on the node itself, what about this?

    public class Node<T>
    {
        public T Data;
        public Node<T> Left;
        public Node<T> Right;

        public static void ForEach(Node<T> root, Action<T> action)
        {
            action(root.Data);

            if (root.Left != null)
                ForEach(root.Left, action);
            if (root.Right != null)
                ForEach(root.Right, action);
        }
    }

    public interface IHasSize
    {
        long Size { get; }
    }

    public static long SumSize<T>(Node<T> root) where T : IHasSize
    {
        long sum = 0;
        Node<T>.ForEach(root, delegate(T item)
        {
            sum += item.Size;
        });
        return sum;
    }

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