<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 Solution:
    def lexicalOrder(self, n: int) -> List[int]:
        def dfs(curr):
            nonlocal ans
            if int(''.join(curr)) <= n:
                ans.append(int(''.join(curr)))
            for d in range(10):
                if int(curr + str(d)) <= n:
                    dfs(curr +  str(d))
        ans = []
        for d in range(1, 10):
            dfs(str(d))
        return ans


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