[LeetCode 208] 实现 Trie (前缀树)

208. 实现 Trie (前缀树)

字典树

转自评论中某大佬，加入了一个preorder可以查看一下树的构造是不是对的

注意：search和startsWith只差了最后一句return。
因为insert的时候插入的是word，最后一个字符结点会保留isWord=true，但是中间的就不一定了，所以startsWith直接返回true，search返回t->isWord

#include 
#include 
#include 
using namespace std;

class Trie {
    Trie *child[26];
    bool isWord;

  public:
    /** Initialize your data structure here. */
    Trie() {
        isWord = false;
        for (int i = 0; i < 26; i++) {
            child[i] = nullptr;
        }
    }

    void preorder(Trie *node) {
        if (node == nullptr)
            return;
        for (int i = 0; i < 26; i++) {
            if (node->child[i] != nullptr) {
                cout << char(i + 'a') << " ";
                preorder(node->child[i]);
            }
        }
    }

    /** Inserts a word into the trie. */
    void insert(string word) {
        Trie *t = this;
        for (char c : word) {
            if (!t->child[c - 'a']) {
                t->child[c - 'a'] = new Trie();
            }
            t = t->child[c - 'a'];
        }
        t->isWord = true;
    }

    /** Returns if the word is in the trie. */
    bool search(string word) {
        Trie *t = this;
        for (char c : word) {
            if (!t->child[c - 'a']) {
                return false;
            }
            t = t->child[c - 'a'];
        }
        return t->isWord;
    }

    /** Returns if there is any word in the trie that starts with the given
     * prefix. */
    bool startsWith(string prefix) {
        Trie *t = this;
        for (char c : prefix) {
            if (!t->child[c - 'a']) {
                return false;
            }
            t = t->child[c - 'a'];
        }
        return true;
    }
};

/**
 * Your Trie object will be instantiated and called as such:
 * Trie* obj = new Trie();
 * obj->insert(word);
 * bool param_2 = obj->search(word);
 * bool param_3 = obj->startsWith(prefix);
 */

int main() {
    Trie s;
    s.insert(string("asdf"));
    s.insert(string("aszx"));
    s.insert(string("abss"));
    s.preorder(&s);

    return 0;
}