使用Java Trie时，无法识别词尾。在递归中失败

Question

我正试图创建我自己的JavaTrie版本，以便拥有它，并获得制作它所需的知识，但这个项目让我感到困惑。我这里有一个非常基本的坏球

我在Trie中添加了3个单词（使用单词的第一个字符作为键，值是额外的三元组）。如果您注意到了，我在trinode类中有print语句，用于在程序运行时检查isWord的值。我希望单词中的最后一个字符将isWord变量设置为True。从而识别一个单词的结尾（稍后我将使用它来获取整个单词）

当我第一次输入这三个单词时，输出会打印每个单词进入Trie时的每个字母，并正确识别哪些字符是单词的结尾

但是，如果在输入单词后立即遍历Trie并重新打印Trie中的所有字符及其isWord状态，“Hello”中的“e”现在突然被标识为单词结尾

我已经滔滔不绝地讲了好几个小时，但我不明白为什么会发生这种情况。以下是工作代码：

package testcode;


import java.util.*;

public class TestCode {

    public static Trie t;
    public static void main (String[] args){
        t = new Trie();
        t.addWord("hello");
        t.addWord("hi");
        t.addWord("soup");
        //at this point the output correctly identifies word endings.
        t.findWords();
        /* but when iterating through the hash map it becomes evident that
        * when entering the word 'hi' the 'e' in 'hello' had its isWord variable
        * changed to true. I followed the logic and I do not see how or why this
        * is happening.
        */
    }
}

//This Trie class handles the root trie, and Trie commands.
class Trie{
    private TrieNode root;

    public Trie(){
        root = new TrieNode();
    }

    public void addWord(String word){
        root.addWord(word.toLowerCase());
    }

    public void findWords(){
        root.findWords();
    }
}

//Trie Node handles the nodes and words within the trie
class TrieNode{

    private TrieNode parent;
    private boolean isWord;
    private boolean hasChildren;
    private char character;
    private Map<Character, TrieNode> children = new HashMap<>();

    public TrieNode(){

        hasChildren = false;
        isWord = false;
    }

    public TrieNode(String word){

        this();
        addWord(word);

    }
    public void addWord(String word){

       char firstChar = word.charAt(0);


       if (children.get(firstChar) == null){

           if(word.length() > 1){

               hasChildren = true;
               children.put(firstChar, new TrieNode(word.substring(1)));
               children.get(firstChar).parent = this;
               System.out.print(firstChar + "--");
               System.out.println(isWord);
           }

           else{
               children.put(firstChar, new TrieNode());
               if(character == 'e'){
                   System.out.println("shits about to go down");
               }
               isWord = true;
               System.out.print(firstChar + "--");
               System.out.println(isWord);
           }
           children.get(firstChar).character = firstChar;
       }

       else {
           children.get(firstChar).addWord(word.substring(1));
       }
   }

    public void findWords(){
        for(Character key : children.keySet()){
            children.get(key).findWords();
            System.out.println(children.get(key).character + " -- " + isWord);     
      }
    }
}

此代码生成以下输出：

o--true
l--false
l--false
e--false
h--false
i--true
p--true
u--false
o--false
s--false
p -- true
u -- false
o -- false
s -- false
o -- true
l -- false
l -- false
e -- true    //notice the e here is now suddenly a word ending with isWord = true
i -- true
h -- false

Answer 1

有一系列可能的问题。。父/子混淆，在父节点处理叶子案例，包括在建筑和；打印输出等

我注意到，在旧的“findWords”代码中，您打印的是子字符，但父“isWord”标志。构建trie时，“存在子节点”和“创建子节点路径”之间存在不必要的差异，因此“isWord”只能标记在新路径上，而不能标记在现有路径上。构建trie似乎也会在父节点而不是叶节点上设置“isWord”

一般来说，如果案例很可能不可靠，那么代码就是嵌套的意大利面条。在可能的情况下，代码应该是通用的，将其保留在方法的主流程中，除非它真的属于IF

这是clean&；正确代码：

class TrieNode{
    private TrieNode parent;
    private boolean isWord;
    private boolean hasChildren;
    private char character;
    private Map<Character, TrieNode> children = new HashMap<>();

    public TrieNode(){
        this.hasChildren = false;
        this.isWord = false;
    }
    public TrieNode (char ch) {
        this.character = ch;
        this.hasChildren = false;
        this.isWord = false;
    }

    public void addWord (String word){
        if (word.length() == 0) {
            this.isWord = true;
            System.out.println( character + "   " + isWord);
            return;
        }

        // represent the Child Node;
        //        
        char firstChar = word.charAt(0);
        TrieNode child = children.get( firstChar);
        if (child == null){
            child = new TrieNode( firstChar);
            children.put( firstChar, child);
            child.parent = this;
            hasChildren = true;
        }

        // add Remaining Word;
        //        call for 1-length words, as 0-length at Child sets 'IsWord'!
        child.addWord( word.substring(1));

        // print building here.
        System.out.println( character + "   " + isWord);
    }



    public void findWords(){
        for(Character key : children.keySet()){
            children.get(key).findWords();
        }
        System.out.println( character + "   " + isWord);     
    }
}