<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 Solution {
private:
    struct Trie{
        vector<Trie*> next;
        string word;
        Trie() : next(vector<Trie*>(26,nullptr)), word("")  {};
    };
public:
    string longestWord(vector<string>& words) {
        Trie root;
        for (auto s: words) {
            Trie* node = &root;
            for (char c: s) {
                int idx = c -'a';
                if (node->next[idx] ==nullptr) {
                    node->next[idx] = new Trie;
                } 
                node = node->next[idx];
            }
            node->word = s;
        }
        string ans;
        for (int i=0;i< 26;i++) {
            dfs(root.next[i],ans);
        }
        return ans;
    }
    
    void dfs(Trie* node, string& ans) {
        if (node ==nullptr) return;
        if (node->word == "") return;
        if (node->word.size() > ans.size()) ans = node->word;
        for (int i=0;i< 26;i++) {
            dfs(node->next[i],ans);
        }
    }
};
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