<p>Given an array of strings <code>words</code> representing an English Dictionary, return <em>the longest word in</em> <code>words</code> <em>that can be built one character at a time by other words in</em> <code>words</code>.</p>

<p>If there is more than one possible answer, return the longest word with the smallest lexicographical order. If there is no answer, return the empty string.</p>

<p>Note that the word should be built from left to right with each additional character being added to the end of a previous word.&nbsp;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;w&quot;,&quot;wo&quot;,&quot;wor&quot;,&quot;worl&quot;,&quot;world&quot;]
<strong>Output:</strong> &quot;world&quot;
<strong>Explanation:</strong> The word &quot;world&quot; can be built one character at a time by &quot;w&quot;, &quot;wo&quot;, &quot;wor&quot;, and &quot;worl&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;banana&quot;,&quot;app&quot;,&quot;appl&quot;,&quot;ap&quot;,&quot;apply&quot;,&quot;apple&quot;]
<strong>Output:</strong> &quot;apple&quot;
<strong>Explanation:</strong> Both &quot;apply&quot; and &quot;apple&quot; can be built from other words in the dictionary. However, &quot;apple&quot; is lexicographically smaller than &quot;apply&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>

class TrieNode:
    def __init__(self, letter: chr = None):
        self.letter = letter
        self.next_letters = [None for _ in range(26)]
        self.is_word = False
    
    def addWord(self, word: str) -> None:
        node = self
        for i, c in enumerate(word):
            next_i = ord(c) - ord('a')
            if node.next_letters[next_i] is None:
                node.next_letters[next_i] = TrieNode(c)
            node = node.next_letters[next_i]
        
        node.is_word = True

class Solution:
    def longestWord(self, words: List[str]) -> str:
        root = TrieNode()
        for word in words:
            root.addWord(word)
        
        self.longest_word = ""
        self.findLongest(root, "")
        return self.longest_word
        
        
    def findLongest(self, node: TrieNode, previous_word: str) -> None:
        for i, next_letter in enumerate(node.next_letters):
            if next_letter is not None and next_letter.is_word:
                new_word = previous_word + chr(i + ord('a'))
                if len(new_word) > len(self.longest_word) or (len(new_word) == len(self.longest_word) and new_word < self.longest_word):
                    self.longest_word = new_word
                self.findLongest(next_letter, new_word)


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