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.

click to show hint.

You should be familiar with how a Trie works. If not, please work on this problem:  Implement Trie (Prefix Tree) first.

思路：这道题如果做过之前的那道
Implement Trie (Prefix Tree) 实现字典树(前缀树)的话就没有太大的难度了，因为这道题里面'.'可以代替任意字符，所以一旦有了'.'，就需要查找之前存下的所有下一层的不是null的path；String match 的题，一般都是DFS 参数里面加入index，然后递归 subproblem求解；

class WordDictionary {
    class TrieNode {
        public TrieNode[] children;
        public boolean isword;
        public String word;
        public TrieNode() {
            this.children = new TrieNode[26];
            this.isword = false;
            this.word = null;
        }
    }

     class Trie {
        public TrieNode root;
        public Trie() {
           this.root = new TrieNode();
        }
        
        public void insert(String word) {
            TrieNode cur = root;
            for(int i = 0; i < word.length(); i++) {
                char c = word.charAt(i);
                if(cur.children[c - 'a'] == null) {
                    cur.children[c - 'a'] = new TrieNode();
                }
                cur = cur.children[c - 'a'];
            }
            cur.isword = true;
            cur.word = word;
        }
    
        public boolean search(String word) {
            return searchHelper(word, root, 0);
        }
         
        private boolean searchHelper(String word, TrieNode cur, int index) {
            if(index == word.length()) {
                return cur.isword;
            }
            char c = word.charAt(index);
            if(c == '.') {
                for(int i = 0; i < 26; i++) {
                    if(cur.children[i] != null) {
                        if(searchHelper(word, cur.children[i], index + 1)) {
                            return true;
                        }
                    }
                }
                return false;
            } else {
                if(cur.children[c - 'a'] == null) {
                    return false;
                } else {
                    return searchHelper(word, cur.children[c - 'a'], index + 1);
                }
            }
        }
     }
    
    private Trie trie;
    public WordDictionary() {
        this.trie = new Trie();
    }
    
    public void addWord(String word) {
        trie.insert(word);
    }
    
    public boolean search(String word) {
        return trie.search(word);
    }
}

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