Numbers With Same Consecutive Differences
<p>Given two integers n and k, return <em>an array of all the integers of length </em><code>n</code><em> where the difference between every two consecutive digits is </em><code>k</code>. You may return the answer in <strong>any order</strong>.</p>

<p>Note that the integers should not have leading zeros. Integers as <code>02</code> and <code>043</code> are not allowed.</p>

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

<pre>
<strong>Input:</strong> n = 3, k = 7
<strong>Output:</strong> [181,292,707,818,929]
<strong>Explanation:</strong> Note that 070 is not a valid number, because it has leading zeroes.
</pre>

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

<pre>
<strong>Input:</strong> n = 2, k = 1
<strong>Output:</strong> [10,12,21,23,32,34,43,45,54,56,65,67,76,78,87,89,98]
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 9</code></li>
	<li><code>0 &lt;= k &lt;= 9</code></li>
</ul>
Breadth-First Search; Backtracking
class Solution:
    def numsSameConsecDiff(self, n: int, k: int) -> List[int]:
        def dfs(x: int):
            if x >= boundary:
                ans.append(x)
                return
            last = x % 10
            if last + k <= 9:
                dfs(x * 10 + last + k)
            if last - k >= 0 and k != 0:
                dfs(x * 10 + last - k)

        ans = []
        boundary = 10 ** (n - 1)
        for i in range(1, 10):
            dfs(i)
        return ans
