Leetcode 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

思路：
如果去掉了特殊字符.就是一道典型的trie树解法题目，所以关键是如何解决特殊字符.的情况。
自己第一次想的是，每次add一个字符的时候同样add一个.字符，即把.当做第27个字符来看待，但是增加第一个.字符之后，如何解决后续字符在.节点的分支是个问题，同样在搜索的方法实现上也很复杂，自己没有想清楚怎么解决。
因此另一种实现方法，就是暴力的，如果遇见了.，则在当前节点的所有子节点进行搜索，如果字符串很长，并且trie树很茂盛，时间复杂度会很高，但是被accepted了！

class WordDictionary {

class TrieNode {
    public boolean isWord;
    public TrieNode[] children;
    public TrieNode() {
        this.isWord = false;
        this.children = new TrieNode[26];
    }
}

TrieNode root;

public WordDictionary() {
    this.root = new TrieNode();
}

/** Adds a word into the data structure. */
public void addWord(String word) {
    TrieNode dummy = this.root;
    for (int i = 0; i < word.length(); i++) {
        char c = word.charAt(i);
        if (dummy.children[c - 'a'] == null) {
            dummy.children[c - 'a'] = new TrieNode();
        }
        dummy = dummy.children[c - 'a'];
    }
    dummy.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) {
    return this._searchHelper(word, 0, this.root);
}

private boolean _searchHelper(String word, int pos, TrieNode node) {
    if (pos == word.length()) {
        return node.isWord;
    }

    char c = word.charAt(pos);
    if (word.charAt(pos) != '.') {
        if (node.children[c - 'a'] == null) {
            return false;
        }
        return _searchHelper(word, pos + 1, node.children[c - 'a']);
    } else {
        for (int i = 0; i < 26; i++) {
            if (node.children[i] != null && _searchHelper(word, pos + 1, node.children[i])) {
                return true;
            }
        }
        return false;
    }
}

}