211. Add and Search Word - Data structure design (Medium)

Design a data structure that supports the following two operations:

void addWord(word)
bool search(word)

search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

For example:

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

Note:

You may assume that all words are consist of lowercase letters a-z.

Solution:

struct TrieNode {
    bool isWord;
    vector<TrieNode*> children;
    TrieNode(): isWord(false), children(26, NULL) {}
};

class WordDictionary {
    bool helper(TrieNode* p, string& word, int idx) {
        if (idx == word.size()) return p->isWord;

        if (word[idx] == '.') {
            for (auto& a: p->children) {
                if (a && helper(a, word, idx+1)) return true;
            }
        } else {
            int pos = word[idx]-'a';
            if (p->children[pos]) return helper(p->children[pos], word, idx+1);
        }
        return false;
    }
public:
    /** Initialize your data structure here. */
    WordDictionary() {
        root = new TrieNode();
    }

    /** Adds a word into the data structure. */
    void addWord(string word) {
        TrieNode* cur = root;
        for (char c: word) {
            int pos = c-'a';
            if (!cur->children[pos]) cur->children[pos] = new TrieNode();
            cur = cur->children[pos];
        }
        cur->isWord = 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 helper(root, word, 0);
    }
private:
    TrieNode* root;
};

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