Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.
Below is one possible representation of s1 = "great":
    great
   /    \
  gr    eat
 / \    /  \
g   r  e   at
           / \
          a   t
To scramble the string, we may choose any non-leaf node and swap its two children.
For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".
    rgeat
   /    \
  rg    eat
 / \    /  \
r   g  e   at
           / \
          a   t
We say that "rgeat" is a scrambled string of "great".
Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".
    rgtae
   /    \
  rg    tae
 / \    /  \
r   g  ta  e
       / \
      t   a
We say that "rgtae" is a scrambled string of "great".
Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.
Example 1:
Input: s1 = "great", s2 = "rgeat"
Output: true
Example 2:
Input: s1 = "abcde", s2 = "caebd"
Output: false
class Solution:
     def isScramble(self, A, B):
         if len(A) != len(B) or sorted(A) != sorted(B):
             return False
 
         if len(A) == 1 or A == B:
             return True
 
         for i in range(1, len(A)):
             if self.isScramble(A[:i], B[:i]) and self.isScramble(A[i:], B[i:]):
                 return True
             elif self.isScramble(A[:i], B[-i:]) and self.isScramble(A[i:], B[:-i]):
                 return True
         return False
class Solution:
     def isScramble(self, A, B):
         if len(A) != len(B) or sorted(A) != sorted(B):
             return False
 
         if len(A) == 1 or A == B:
             return True
 
         for i in range(1, len(A)):
             if self.isScramble(A[:i], B[:i]) and self.isScramble(A[i:], B[i:]):
                 return True
             elif self.isScramble(A[:i], B[-i:]) and self.isScramble(A[i:], B[:-i]):
                 return True
         return False