211. Add and Search Word - Data structure design

https://leetcode.com/problems/add-and-search-word-data-structure-design/

Solution:
May use the prefix trie data structure. Then need DSF for the word searching. As long as there is one path is right, return true; otherwise, return false.

Code:

class WordDictionary {
public:
    struct TrieNode{
        bool end;
        vector<TrieNode*> children;
        TrieNode(): end(false), children(26, NULL) {}
    };
    
    WordDictionary(){//constructor;
        root = new TrieNode();
    }
    // Adds a word into the data structure.
    void addWord(string word) {
        TrieNode *t = root;
        for(auto a:word){
            if(!t->children[a-'a']) t->children[a-'a'] = new TrieNode();
            t = t->children[a-'a'];
        }
        t->end = true;
    }

    // Returns if the word is in the data structure. A word could
    // contain the dot character '.' to represent any one letter.
    bool search(string word) {
        return swDSF(word, root, 0);
    }
    
    bool swDSF(string w, TrieNode* r, int i){
        if(i == w.size()) return r->end;
        if(w[i] == '.'){
             for(auto a:r->children){
                if(a && swDSF(w, a, i+1)) return true;
            }
            return false;
        }
        return r->children[w[i]-'a'] && swDSF(w, r->children[w[i]-'a'], i+1);
    }
    
private:
    TrieNode *root;
};
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");