Add and Search Word - Data Structure Design

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.

Analyse: The same as Implement Trie. If '.' is encountered, we need to traversal all situations. 

Runtime: 88ms.

class TrieNode{
public:
    TrieNode* childNode[27];
    bool isVal;
    TrieNode(){
        isVal = false;
        for(int i = 0; i < 27; i++) 
            childNode[i] = NULL;
    }
};
class WordDictionary {
private: 
    TrieNode* root;
public:
    WordDictionary(){
        root = new TrieNode();
    }
    // Adds a word into the data structure.
    void addWord(string word) {
        TrieNode* temp = root;
        for(int i = 0; i < word.size(); i++){
            int index = word[i] - 'a';
            if(word[i] == '.') 
                index = 26;
            else{
                if(temp->childNode[index] == NULL)
                    temp->childNode[index] = new TrieNode();
            }
            temp = temp->childNode[index];
        }
        temp->isVal = true; //word has been inserted
    }

    // 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 levelSearch(word, root, 0);
    }
    
    bool levelSearch(string& word, TrieNode* root, int index){
        TrieNode* temp = root;
        if(!temp) return false; //length of the word is larger than that of the strings in the trie
        if(index >= word.size()) return temp->isVal;
        
        if(word[index] != '.') 
            return levelSearch(word, temp->childNode[word[index] - 'a'], index + 1);
        else{
            for(int i = 0; i < 26; i++)
                if(levelSearch(word, temp->childNode[i], index + 1)) return true;
            return false;
        }
    }
};

// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");