# 208. Implement Trie (Prefix Tree)

**<font color=orange>难度: Medium</font>**

## 刷题内容

> 原题连接

* https://leetcode.com/problems/implement-trie-prefix-tree/

> 内容描述

```

Implement a trie with insert, search, and startsWith methods.

Example:

Trie trie = new Trie();

trie.insert("apple");
trie.search("apple");   // returns true
trie.search("app");     // returns false
trie.startsWith("app"); // returns true
trie.insert("app");   
trie.search("app");     // returns true
Note:

You may assume that all inputs are consist of lowercase letters a-z.
All inputs are guaranteed to be non-empty strings.
```

## 解题方案

> 思路 1
******- 时间复杂度: O(N)******- 空间复杂度: O(N)******

这个Python实现也太精美了吧，谷歌复写之

然后还unlock了一个solution，to read

Trie整个都需要 to read，精美，可爱😊




```python
class TrieNode(object):
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.childs = dict()
        self.isWord = False
        
        

class Trie(object):

    def __init__(self):
        self.root = TrieNode()

    def insert(self, word):
        """
        Inserts a word into the trie.
        :type word: str
        :rtype: void
        """
        node = self.root
        for letter in word:
            child = node.childs.get(letter)
            if child is None:
                child = TrieNode()
                node.childs[letter] = child
            node = child
        node.isWord = True

    def search(self, word):
        """
        Returns if the word is in the trie.
        :type word: str
        :rtype: bool
        """
        node = self.root
        for i in word:
            child = node.childs.get(i)
            if child is None:
                return False
            node = child
        return node.isWord
        

    def startsWith(self, prefix):
        """
        Returns if there is any word in the trie
        that starts with the given prefix.
        :type prefix: str
        :rtype: bool
        """
        node = self.root
        for letter in prefix:
            child = node.childs.get(letter)
            if child is None:
                return False
            node = child
        return True
        

# Your Trie object will be instantiated and called as such:
# trie = Trie()
# trie.insert("somestring")
# trie.search("key")
        
```