208.Implement Trie (Prefix Tree) & 211. Add and Search Word - Data structure design

class TrieNode{
public:
    TrieNode* son[26];
    bool isWord;
    TrieNode():isWord(false){
        for(auto &e:son)
            e=nullptr;
    }
};
class Trie {
public:
    /** Initialize your data structure here. */
    Trie():root(new TrieNode()) {}
    
    /** Inserts a word into the trie. */
    void insert(const string& word) {
        auto p=root;
        for(auto c : word){
            if(!p->son[c-'a'])p->son[c-'a']=new TrieNode();
            p=p->son[c-'a'];
        }
        p->isWord=true;
    }
    
    /** Returns if the word is in the trie. */
    bool search(const string& word) {
        auto p=root;
        for(auto c : word){
            if(!p->son[c-'a'])return false;
            p=p->son[c-'a'];
        }
        return p->isWord;
    }
    
    /** Returns if there is any word in the trie that starts with the given prefix. */
    bool startsWith(const string& prefix) {
        auto p=root;
        for(auto c : prefix){
            if(!p->son[c-'a'])return false;
            p=p->son[c-'a'];
        }
        return true;
    }
private:
    TrieNode* root;
};


class Tries{
public:
    Tries():isKey(false),son(26,nullptr)
    {}

    vector<Tries*> son;
    bool isKey;
};
class WordDictionary {
public:
    /** Initialize your data structure here. */
    WordDictionary():root(new Tries()) {}
    
    /** Adds a word into the data structure. */
    void addWord(const string& word) {
        auto p=root;
        for(auto c : word){
            if(p){
                if(!p->son[c-'a'])
                    p->son[c-'a']=new Tries();
                p=p->son[c-'a'];
            }
        }
        p->isKey=true;
    }
    
    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    bool search(const string& word) {
        if(word.empty())return false;
        return query(word, root);
    }
private:
    Tries*root;
    
    bool query(const string& word, Tries* r){
        auto p=r;
        auto n=word.size();
        for(int i=0; i<n; ++i){
            if(p && word[i]!='.')
                p=p->son[word[i]-'a'];
            else if(p && word[i]=='.'){
                auto tmp=p;
                for(auto ele : p->son){
                    p=ele;
                    if(query(word.substr(i+1), ele))
                        return true;
                }
                
            }
            else break;
        }
        return p && p->isKey;
    }
};

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * bool param_2 = obj.search(word);
 */