LeetCode: 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

.

class TrieNode {
public:
// Initialize your data structure here.
TrieNode(): child({NULL}), isEnd(false)
{

}
bool isEnd;
TrieNode *child[26];
};
class WordDictionary {
public:

// Adds a word into the data structure.
WordDictionary()
{
root = new TrieNode();

}

void addWord(string word) {
TrieNode *iter = root;
for(int i = 0; i < word.size(); i++)
{
if(iter->child[word[i] - 'a'] != NULL)
{
iter = iter->child[word[i] - 'a'];
}
else
{
TrieNode *temp = new TrieNode();
iter->child[word[i] - 'a'] = temp;
iter = temp;
}
}
iter->isEnd = 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) {
TrieNode *iter = root;
return dfs(word, iter);
}
private:
bool dfs(string &word, TrieNode *iter)
{
if(word.empty())
return iter->isEnd;
if('.' == word[0])
{
bool result = false;
for(int j = 0; j < 26; j++)
{
if(NULL == iter->child[j])
continue;
string sub = word.size() <= 1 ? "" : word.substr(1);
result |= dfs(sub, iter->child[j]);
if(true == result)
return true;
}
return false;
}
else if(iter->child[word[0] - 'a'] != NULL)
{
string sub = word.size() <= 1 ? "" : word.substr(1);
return dfs(sub, iter->child[word[0] - 'a']);
}
else
{
return false;
}

}
TrieNode* root;
};