Question

根据我对三元搜索树的理解，它们在可以寻找和找到的项目中是反向确定的（不确定正确的术语）。我的意思是，如果你为 cat ，自行车，轴创建一个三元树，你给某人三元树，他应该能够从中扣除这三个字。

这是对的吗？

我问，因为我有一个三元树结构，包含像ISMAP，SELECTED和COMPACT（确实是HTML 4的属性）这样的单词，我想知道是否可以获得存储在该树中的项目的完整列表（原始文件消失了）。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, 
};

Answer 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，用

启动它

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，我认为就是这样。

好的，这里有一些基于简单数组表示的C＃代码。我使用了这个维基百科文章中的树

http://en.wikipedia.org/wiki/Ternary_search_tree

我把它表示为一个数组，其中根是元素0，然后它的孩子是1,2,3。1的孩子是4,5,6，依此类推。 '\ 0'用于表示没有更多的孩子。算法与上述相同。

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, "");
        }
    }
}

打印

cute
cup
at
as
he
us
i

Answer 2

数据结构并不是三元树，因为第三个分支是隐式的（即，是当前条目之后的下一个条目）。它有点像在二叉树结构中实现的trie。每4个数字对应一个类似struct { char letter, Loff, Roff, flag}的结构。例如，条目0 = 72,5,77,0是字母'H'，左偏移5，右偏移77，标志0（可能意味着不是终端）。在左偏移后，在＃0之后的5个条目我们有67,12,40,0，即C, 12, 40, 0; ＃5之后的12个条目，65,0,0,0为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ACTION。在右偏移之后，＃0之后的77个条目我们有78,8,48,0, 79,36,0,0, 83,30,55,0, 72,0,0,0, 65,...或带有分支的N，O和S条目，后跟没有显式分支的H，A，D，E条目，以生成NOSHADE。

当你沿着树向树叶方向移动时，在当前字符串中添加字母（如在trie中遍历），当你向上（远离树叶）时，从当前字符串的末尾删除字母。

是否有可能在三元搜索树中生成所有可能的术语？

2 个答案: