<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, char=""):
        self.char = char
        self.isWord = False
        self.children = [None] * 26

class Solution:
    def __init__(self):
        self.root = TrieNode()
        self.root.isWord = True
    
    def addWord(self, w):
        curr = self.root
        for c in w:
            index = ord(c) - ord('a')
            if not curr.children[index]:
                curr.children[index] = TrieNode(c)
            curr = curr.children[index]
        curr.isWord = True
    
    def dfsTrie(self, curr):
        if not curr or not curr.isWord:
            return ""
        res = ""
        mxLen = 0
        for i in range(26):
            dfs = self.dfsTrie(curr.children[i])
            if len(dfs) > mxLen:
                res = dfs
                mxLen = len(dfs)
        return curr.char + res
        
    def longestWord(self, words: List[str]) -> str:
        # add each word to Trie
        for w in words:
            self.addWord(w)
        
        # walk Trie and find longest height on isWord == True streak
        return self.dfsTrie(self.root)
18100