<p>Given an integer <code>n</code>, return all the numbers in the range <code>[1, n]</code> sorted in lexicographical order.</p>

<p>You must write an algorithm that runs in&nbsp;<code>O(n)</code>&nbsp;time and uses <code>O(1)</code> extra space.&nbsp;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 13
<strong>Output:</strong> [1,10,11,12,13,2,3,4,5,6,7,8,9]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 2
<strong>Output:</strong> [1,2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
</ul>

class TrieNode:
    def __init__(self):
        self.children = {}
        self.endOfWord = False
    
    def insert(self, word):
        cur = self

        for c in word:
            if c not in cur.children:
                cur.children[c] = TrieNode()
            cur = cur.children[c]
        cur.endOfWord = True

class Solution:
    def lexicalOrder(self, n: int) -> List[int]:
        
        self.trie = TrieNode()
        for i in range(1, n+1):
            self.trie.insert(str(i))
            
        def dfs(node, cur, result):
            if node.endOfWord:
                result.append(int(''.join(cur)))
            for key in node.children.keys():
                cur.append(key)
                dfs(node.children[key], cur, result)
                cur.pop()
            


        result = []
        dfs(self.trie, [], result)
        # print(result)

        return result

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