Question

我正在尝试通过B-Tree实现它来构建c ++中的trie。每个节点都有一个键，一个指向其他节点的指针数组，以及一个布尔值，用于确定它是否为叶子。默认情况下，初始化新节点时所有指针都指向null，每当添加新节点时，前一个节点都指向新节点。

但由于某种原因，我的其他if（currentNode.children [arrayPointer]！= NULL）永远不会被调用。我完全难过了。我不确定是不是因为我标注了错误或者某些事情的顺序不正确。

#include <iostream>
using namespace std;

numberOfWords = 0;
numberOfNodes = 0;

struct Node;
struct Node;
    struct Node {
    char key;
    bool isLeaf;
    struct Node *children[36]; //26 letters + 10 digits
};



main() {
    //initializing the root
    Node root;
    root.key ='$';
    int i = 0;
    while (i < 36) {
        root.children[i] = NULL;    
        i++;
    }
    root.isLeaf = 1;
    numberOfNodes++;

    insertWord("the", root);
    insertWord("and", root);
    insertWord("there", root);

    cout << numberOfWords << " words found." << endl;
    cout << numberOfNodes << " nodes found." << endl;

}

void insertWord(string word, Node currentNode) {
    int wordLength = word.length();
    char letterToInsert;
    letterToInsert = word[0];
    int arrayPointer;
    arrayPointer = charToInteger(letterToInsert);


    //if the node is not found, a new path is inserted
    if (currentNode.children[arrayPointer] == NULL) {
        insertNewPath(word, currentNode);
    }

    //if the node with the current letter is found, I want to recursively call
    //insertWord and pass in the updated word and the next node.    

    //For some reason, this statement is never being called.
    else if (currentNode.children[arrayPointer] != NULL)  {
        string updatedWord;
        updatedWord = word.substr(1,wordLength);
        Node nextNode;
        currentNode.children[arrayPointer] = &nextNode;
        insertWord(updatedWord, nextNode);
    }

}       

//This function will keep recursively calling itself and passing in the new word 
//and the new node until the each letter is added.
void insertNewPath(string word, Node currentNode) {
    int wordLength;
    wordLength = word.length();
    if (wordLength > 0) {
        char letterToInsert;
        letterToInsert = word[0];
        int arrayPointer;
        //charToInteger takes a character and converts it into an integer value - a=0,b=1,...,8=34,9=35
        arrayPointer = charToInteger(letterToInsert);
        Node newNode;
        numberOfNodes++;
        newNode.key = letterToInsert;
        // Here, I set the current node's pointer to point to the new node. 
        // This should cause line 168 to execute when entering an existing
        // letter to the trie but for some reason it never does.
        currentNode.children[arrayPointer] = &newNode;
        //setting every pointer in the new node to null
        int i = 0;
        while (i < 36) {
            newNode.children[i] = NULL; 
            i++;
        }
        currentNode.isLeaf = 0;
        string updatedWord;
        updatedWord = word.substr(1, wordLength);
        insertNewPath(updatedWord, newNode);
    }
    else if (wordLength == 0){
        currentNode.isLeaf = 1;
        //Once all the letters have been added, the last 
        //node/letter to be added is set as a leaf, making
        //it a complete word.
    }   
}

我期望得到的输出是

3 words found.
9 nodes found.

相反，我正在

3 words found.
11 nodes found.

因此，为单个字母创建一个节点。

c ++ trie - 即使在插入

0 个答案: