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);        }    }}