<p>An <strong>n-bit gray code sequence</strong> is a sequence of <code>2<sup>n</sup></code> integers where:</p>

<ul>
	<li>Every integer is in the <strong>inclusive</strong> range <code>[0, 2<sup>n</sup> - 1]</code>,</li>
	<li>The first integer is <code>0</code>,</li>
	<li>An integer appears <strong>no more than once</strong> in the sequence,</li>
	<li>The binary representation of every pair of <strong>adjacent</strong> integers differs by <strong>exactly one bit</strong>, and</li>
	<li>The binary representation of the <strong>first</strong> and <strong>last</strong> integers differs by <strong>exactly one bit</strong>.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>any valid <strong>n-bit gray code sequence</strong></em>.</p>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> [0,1,3,2]
<strong>Explanation:</strong>
The binary representation of [0,1,3,2] is [00,01,11,10].
- 0<u>0</u> and 0<u>1</u> differ by one bit
- <u>0</u>1 and <u>1</u>1 differ by one bit
- 1<u>1</u> and 1<u>0</u> differ by one bit
- <u>1</u>0 and <u>0</u>0 differ by one bit
[0,2,3,1] is also a valid gray code sequence, whose binary representation is [00,10,11,01].
- <u>0</u>0 and <u>1</u>0 differ by one bit
- 1<u>0</u> and 1<u>1</u> differ by one bit
- <u>1</u>1 and <u>0</u>1 differ by one bit
- 0<u>1</u> and 0<u>0</u> differ by one bit
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> [0,1]
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 16</code></li>
</ul>

class Solution:

    def __init__(self):
        self.res = []
        self.path = []
        self.used = set()

    def grayCode(self, n: int) -> List[int]:
        if n < 0: return self.res
        self.dfs(n, 0)
        return self.res[0]
    
    def dfs(self, n: int, num:int):
        if self.res:                    # there is one valid res, return
            return
        
        if len(self.path) == (1 << n):  # path has 2^n element, found a res
            self.res.append(self.path[:])
            return
        
        if num in self.used:        # num visited
            return

        self.path.append(num)       # dfs, add node now
        self.used.add(num)

        for i in range(n):
            onediff = self.flipbit(num, i)
            self.dfs(n, onediff)

        self.path.pop()             # dfs, revert node now
        self.used.remove(num)

    def flipbit(self, num:int, pos: int) -> int:
        return num ^ (1 << pos) 

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