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

Note:
You may assume that all words are consist of lowercase letters a-z.

 

public class WordDictionary {
    public class TrieNode
    {
        public Dictionary<char, TrieNode> children = new Dictionary<char, TrieNode>();
        public bool isWord = false;
    }
    
    private TrieNode root = new TrieNode();

    /** Initialize your data structure here. */
    public WordDictionary() {
    }
    
    /** Adds a word into the data structure. */
    public void AddWord(string word) {
        var node = root;
        
        for (int i = 0; i < word.Length; i++)
        {
            if (!node.children.ContainsKey(word[i]))
            {
                node.children[word[i]] = new TrieNode();
            }
            
            node = node.children[word[i]];
        }
        
        node.isWord = true;
    }
    
    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
    public bool Search(string word) {
        return InternalSearch(word, root);
    }
    
    private bool InternalSearch(string word, TrieNode node) {
        for (int i = 0; i < word.Length; i++)
        {
            if (word[i] == '.')
            {
                foreach (var c in node.children.Values)
                {
                    if (InternalSearch(word.Substring(i + 1, word.Length - i - 1), c))
                    {
                        return true;
                    }
                }
                
                return false;
            }
            else if (node.children.ContainsKey(word[i]))
            {
                node = node.children[word[i]];
            }
            else
            {
                return false;
            }
        }
        
        return node.isWord;
    }
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.AddWord(word);
 * bool param_2 = obj.Search(word);
 */