represent any one letter. bool search(string word) { return searchWord(word, root_, 0); } bool searchWord(string word, TrieNode *node, int s) { if (s == word.length()) { return node->isString; } // Match the char. if (node->leaves.find(word[s]) != node->leaves.end()) { return searchWord(word, node->leaves[word[s]], s + 1); } else if (word[s] == '.') { // Skip the char. for (const auto& i : node->leaves) { if (searchWord(word, i.second, s + 1)) { return true; } } } return false; } private: TrieNode *root_; }; // Your WordDictionary object will be instantiated and called as such: // WordDictionary wordDictionary; // wordDictionary.addWord("word"); // wordDictionary.search("pattern"); 4:["$","div",null,{"children":[["$","header",null,{"children":[["$","div",null,{"className":"text-3xl text-bold p-0 mt-2 mb-2","children":["$","h1",null,{"children":["Add and Search Word",["$","span",null,{"className":"font-mono text-sm text-gray-500","children":"(LintCode 473):"}]]}]}],""]}],["$","article",null,{"children":[["$","div",null,{"className":"p-0","dangerouslySetInnerHTML":{"__html":"

Problem description: Visit and Support LintCode Search via
Difficulty: Medium
Tags: LintCode, Stack, O(min(n, h)), O(min(n, h), Medium,

Problem