<p>Given a string <code>s</code>, partition <code>s</code> such that every <span data-keyword="substring-nonempty">substring</span> of the partition is a <span data-keyword="palindrome-string"><strong>palindrome</strong></span>. Return <em>all possible palindrome partitioning of </em><code>s</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "aab"
<strong>Output:</strong> [["a","a","b"],["aa","b"]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "a"
<strong>Output:</strong> [["a"]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 16</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
</ul>

class Solution:
    def isPalindrome(self, s):
        return s == s[::-1]

    def solve(self, str, tempList, ans):
        if len(str) == 0:
            ans.append(tempList[:])  # add a copy
            return

        for i in range(1, len(str) + 1):
            a = str[:i]
            b = str[i:]
            if self.isPalindrome(a):
                tempList.append(a)
                self.solve(b, tempList, ans)
                tempList.pop()  # backtrack

    def partition(self, s: str) -> List[List[str]]:
        ans = []
        tempList = []

        self.solve(s, tempList, ans)

        return ans

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