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：
该题目和”Trie”相似，稍微有些变化。

public class WordDictionary {
    static class Node{
        public Map<Character,Node> map;
        public char c;
        public boolean isTail;
        public Node(char c){
            this.c=c;
            map = new HashMap<>();
            isTail=false;
        }

    }

    private Node root;

    public WordDictionary(){
        root = new Node((char)0);
    }

    // Adds a word into the data structure.
    public void addWord(String word) {
        addWord(word,root);
    }

    private void addWord(String word,Node root){
        if(word.length()==0){
            root.isTail=true;
            return;
        }
        char c=word.charAt(0);
        if(!root.map.containsKey(c)){
            Node n = new Node(c);
            root.map.put(c,n);
        }
        Node n=root.map.get(c);
        addWord(word.substring(1),n);
    }

    // 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 search(word,root);
    }

    private boolean search(String word,Node root){
        if(word.length()==0){
            return root.isTail==true; 
        }
        char c=word.charAt(0);
        if(c==‘.‘){//深度优先搜索
            for(Map.Entry<Character,Node> en:root.map.entrySet()){
                boolean ret = search(word.substring(1),en.getValue());
                if(ret){
                    return true;
                }
            }
            return false;
        }else{
            if(!root.map.containsKey(c)){
                return false;
            }
            Node n=root.map.get(c);
            return search(word.substring(1),n);
        }
    }
}