211. 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

.

similar to 208 , use dfs 

class TrieNode{
boolean isWord;
TrieNode[] children;
public TrieNode(){
children = new TrieNode[26];
isWord = false;
}
}
public class WordDictionary {
TrieNode root;
public WordDictionary(){
root = new TrieNode();
}
// Adds a word into the data structure.
public void addWord(String word) {
TrieNode p=root;
for(int i=0; i<word.length(); i++){
char c = word.charAt(i);
if(p.children[c -'a']==null){
p.children[c -'a']=new TrieNode();
}
p = p.children[c -'a'];
}
p.isWord = true;
}

// Returns if the word is in the data structure. A word could
// contain the dot character '.' to represent any one letter.
public boolean search(String word) {
TrieNode p=root;
return dfs(word, 0, p);
}
public boolean dfs(String word, int i, TrieNode p){
if(i>=word.length()) {
if(p.isWord) return true;
else return false;
}
char c = word.charAt(i);
if(c=='.'){
for(int k=0; k<26; k++){
if(p.children[k]!=null && dfs(word, i+1, p.children[k])) return true;
}
return false;
}else{
if(p.children[c-'a']==null) return false;
return dfs(word, i+1, p.children[c-'a']);
}
}
}

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