C# 是否有可能生成所有可在三元搜索树中查找的可能术语？

从我对三元搜索树的理解来看，它们在可以搜索和找到的项目中是逆确定性的（不确定是否有正确的术语）。我的意思是，如果你为cat，bicycle，axis创建一个三元树，你给某人这个三元树，他应该能够从中扣除这三个词

这是正确的吗

我这样问是因为我有一个三元树结构，其中包含诸如ISMAP、SELECTED和COMPACT（实际上是HTML4的属性）之类的词，我想知道是否可以得到存储在该树中的项目的完整列表（原始文档已经不见了）。结构如下所示：

internal static byte [] htmlAttributes = {
   72,5,77,0, 82,0,0,0, 69,0,0,0, 70,0,0,0, 0,0,0,1, 67,12,40,0, 79,7,0,0,
   77,31,0,0, 80,0,0,0, 65,0,0,0, 67,0,0,0, 84,0,0,0, 0,0,0,2, 73,11,18,0,
   84,0,0,0, 69,0,0,0, 0,0,0,1, 65,0,0,0, 67,0,0,0, 84,0,0,0, 73,0,0,0,
   79,0,0,0, 78,0,0,0, 0,0,0,1, 72,0,0,0, 69,0,0,0, 67,0,0,0, 75,0,0,0,
   69,0,0,0, 68,0,0,0, 0,0,0,2, 76,0,0,0, 65,0,0,0, 83,0,0,0, 83,0,0,0,
   73,0,0,0, 68,0,0,0, 0,0,0,1, 68,0,0,0, 69,0,0,0, 66,0,0,0, 65,0,0,0,
   83,0,0,0, 69,0,0,0, 0,0,0,1, 68,0,28,0, 69,7,15,0, 67,0,22,0, 76,0,0,0,
   65,0,0,0, 82,0,0,0, 69,0,0,0, 0,0,0,2, 65,0,0,0, 84,0,0,0, 65,0,0,0,
   0,0,1,1, 83,0,0,0, 82,0,0,0, 67,0,0,0, 0,0,0,1, 73,0,0,0, 83,0,0,0,
   65,0,0,0, 66,0,0,0, 76,0,0,0, 69,0,0,0, 68,0,0,0, 0,0,0,2, 70,0,0,0,
   69,0,0,0, 82,0,0,0, 0,0,0,2, 70,0,0,0, 79,0,0,0, 82,0,0,0, 0,0,0,1,
   78,8,48,0, 79,36,0,0, 83,30,55,0, 72,0,0,0, 65,0,0,0, 68,0,0,0, 69,0,0,0,
   0,0,0,2, 77,9,0,0, 85,0,0,0, 76,0,0,0, 84,0,0,0, 73,0,0,0, 80,0,0,0,
   76,0,0,0, 69,0,0,0, 0,0,0,2, 73,0,6,0, 83,0,0,0, 77,0,0,0, 65,0,0,0,
   80,0,0,0, 0,0,0,2, 76,0,0,0, 79,0,0,0, 78,0,0,0, 71,0,0,0, 68,0,0,0,
   69,0,0,0, 83,0,0,0, 67,0,0,0, 0,0,0,1, 72,0,9,0, 82,0,0,0, 69,0,0,0,
   70,0,0,0, 0,0,0,2, 65,0,0,0, 77,0,0,0, 69,0,0,0, 0,0,0,1, 82,0,0,0,
   69,0,0,0, 83,0,0,0, 73,0,0,0, 90,0,0,0, 69,0,0,0, 0,0,0,2, 82,14,22,0,
   69,0,0,0, 65,0,0,0, 68,0,0,0, 79,0,0,0, 78,0,0,0, 76,0,0,0, 89,0,0,0,
   0,0,0,2, 87,0,0,0, 82,0,0,0, 65,0,0,0, 80,0,0,0, 0,0,0,2, 80,0,0,0,
   82,0,0,0, 79,0,0,0, 70,0,0,0, 73,0,0,0, 76,0,0,0, 69,0,0,0, 0,0,0,1,
   83,0,12,0, 82,3,0,0, 67,0,0,0, 0,0,0,1, 69,0,0,0, 76,0,0,0, 69,0,0,0,
   67,0,0,0, 84,0,0,0, 69,0,0,0, 68,0,0,0, 0,0,0,2, 85,0,0,0, 83,0,0,0,
   69,0,0,0, 77,0,0,0, 65,0,0,0, 80,0,0,0, 0,0,0,1, 
};

---

我认为算法是这样的

printOutWords(root, wordSoFar)
     if (!root.hasMiddle)
        print wordSoFar + root.char

     if (root.hasMiddle)
        printOutWords(root.middle, wordSoFar + root.char)
     if (root.hasLeft)
        printOutWords(root.left, wordSoFar)
     if (root.hasRight)
        printOutWords(root.right, wordSoFar)

然后，从

printOutWords(ternaryTree, "")

我不知道如何解码你的数组，但如果你能实现这些操作，我想是这样的

好的，下面是一些基于简单数组表示的C代码。我使用了这篇维基百科文章中的树

我将它表示为一个数组，其中根是元素0，然后它的子元素是1、2、3。1的孩子是4岁、5岁、6岁等等\0'用于表示没有更多的孩子。算法同上

using System;
using System.Text;

namespace TreeDecode
{
    class Program
    {
        // http://en.wikipedia.org/wiki/Ternary_search_tree
        //The figure below shows a ternary search tree with the strings "as", "at", "cup", "cute", "he", "i" and "us":
        internal static char[] searchTree = {
                                                                               'c', 
                              'a',                                             'u',                                               'h', 
               '\0',          't',          '\0',            '\0',             't',           '\0',              '\0',            'e',            'u',
         '\0','\0','\0', 's','\0','\0','\0','\0','\0',  '\0','\0','\0',  'p','e','\0',   '\0','\0','\0',    '\0','\0','\0', '\0','\0','\0',   'i','s','\0',
        };

       static void printOutWords(char[] tree, int root, string wordSoFar) {
          if (!HasMiddle(tree, root))
              Console.WriteLine(wordSoFar + CharAt(tree, root));

          if (HasMiddle(tree, root))
              printOutWords(tree, MiddleKid(root), wordSoFar + CharAt(tree, root));
          if (HasLeft(tree, root))
              printOutWords(tree, LeftKid(root), wordSoFar);
          if (HasRight(tree, root))
              printOutWords(tree, RightKid(root), wordSoFar);

        }    

        private static int RightKid(int root)
        {
            return root * 3 + 3;            
        }

        private static bool HasRight(char[] tree, int root)
        {
            int rightIndex = RightKid(root);
            return (rightIndex < tree.Length && tree[rightIndex] != 0);
        }

        private static int LeftKid(int root)
        {
            return root * 3 + 1;
        }

        private static bool HasLeft(char[] tree, int root)
        {
            int leftIndex = LeftKid(root);
            return (leftIndex < tree.Length && tree[leftIndex] != 0);
        }

        private static int MiddleKid(int root)
        {
            return root * 3 + 2;
        }

        private static bool HasMiddle(char[] tree, int root)
        {
            int middleIndex = MiddleKid(root);
            return (middleIndex < tree.Length && tree[middleIndex] != 0);
        }

        private static int NumKids(char[] tree, int root)
        {
            return (HasMiddle(tree, root) ? 1 : 0) + (HasRight(tree, root) ? 1 : 0) + (HasLeft(tree, root) ? 1 : 0);
        }


        private static string CharAt(char[] tree, int root)
        {
            return new String(tree[root], 1);
        }


        static void Main(string[] args)
        {
            printOutWords(searchTree, 0, "");
        }
    }
}

---

数据结构并不完全是三元树，因为第三个分支是隐式的（即当前条目之后的下一个条目）。这有点像是在二叉树结构中实现的。每4个数字对应一个结构，如

struct{char letter，Loff，Roff，flag}

。例如，条目0=

72,5,77,0

是字母“H”，左偏移量为5，右偏移量为77，标志0（可能意味着不是终端）。在左偏移量之后，在#0之后有5个条目，我们有

67,12,40,0

，即

C，12,40,0

；#5，

65,0,0,0

之后的12个条目是

A，0,0,0

。它和接下来的5个条目（65,67,84,73,79,78）显然对应于字符串

动作

。在右偏移量之后，#0之后有77个条目，我们有

78,8,48,0,79,36,0,0,0,83,30,55,0,72,0,0,0,0,65，

或带有分支的N、O和S条目，然后是没有显式分支的H、A、D、E条目，以生成

NOSHADE

---

当您沿着树向叶子方向移动时，将字母添加到当前字符串（如在trie中遍历），当您返回（远离叶子）时，将字母从当前字符串的末尾删除。

谢谢您的帮助。所以至少你确认了我认为是可能的。现在我将尝试实现它。我放入一个示例树和一些C代码来演示如何实现它。我不知道如何读取数组，但是如果你知道的话，这段代码基本上应该可以工作。根据@jwpat7答案中的信息——只需重新实现hasleet、LeftKid等，它也应该可以在你的数组上工作。

cute
cup
at
as
he
us
i