[leedcode 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

public class WordDictionary {
    //字典树的变形，每个节点保存一个字符，以及子节点集合：list，和标记是否是尾节点的isEnd布尔值
    //add和search方法都利用递归进行实现
    private Node root;
    static class Node{
        public Character val;
        public List<Node> children;
        public boolean isEnd;
        public Node(Character val){
            this.val=val;
            children=new ArrayList<Node>();
            isEnd=false;
            
        }
    }
    public WordDictionary(){
        root=new Node('/');
    }

    // Adds a word into the data structure.
    public void addWord(String word) {
        addword(word,root);
    }
    public void addword(String word,Node root){
        if(word.length()==0){
            root.isEnd=true;
            return ;
        }
        Node cur=null;//代表匹配的新节点
        if(root.children.size()>0){
            for(Node node:root.children){
                if(node.val==word.charAt(0)){
                    cur=node;
                    break;
                    
                }
            }
            
        }
        if(cur!=null) addword(word.substring(1),cur);
        else {
            cur =new Node(word.charAt(0));
            root.children.add(cur);
            addword(word.substring(1),cur);
        }
    }

    // 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.isEnd==true;
        }
        if(root.children==null) return false;
        if(word.charAt(0)=='.'){
            for(Node node:root.children){
                boolean ret=search(word.substring(1),node);
                if(ret) return true;
            }
            return false;
        }else{
            for(Node node:root.children){
                if(node.val==word.charAt(0)){
                    return search(word.substring(1),node);
                }
            }
            return false;
        }
        
    }
}

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