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.

class Solution:
    # @return a boolean
    def isScramble(self, s1, s2):
        if len(s1) != len(s2):
            return False
        if s1 == s2:
            return True
            
        length = len(list(s1))
        if sorted(s1) != sorted(s2):
            return False
        
        for i in xrange(1,length):
            if self.isScramble(s1[:i],s2[:i]) and self.isScramble(s1[i:],s2[i:]): 
                return True
            if self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:],s2[:-i]):
               return True 
        return False   

# Note:
# Condition: 1) length_s1 != length_s2
# 2) s1 == s2, s1与s2完全相等
# 3) sorted(s1) 与 sorted(s2)是不是相等
# 4) 比较s1[:i] s2[:i] and s1[i:],s2[i:]
# 5) 比较s1[:i], s2[length_s2-i:] and s1[i:],s2[length_s2:-i]

"""
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.

"""
class Solution:
    # @return a boolean
    def isScramble(self, s1, s2):
        if len(s1) != len(s2):
            return False
        if s1 == s2:
            return True
            
        length = len(list(s1))
        if sorted(s1) != sorted(s2):
            return False
        
        for i in xrange(1,length):
            if self.isScramble(s1[:i],s2[:i]) and self.isScramble(s1[i:],s2[i:]): 
                return True
            if self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:],s2[:-i]):
               return True 
        return False   

# Note:
# Condition: 1) length_s1 != length_s2
# 2) s1 == s2, s1与s2完全相等
# 3) sorted(s1) 与 sorted(s2)是不是相等
# 4) 比较s1[:i] s2[:i] and s1[i:],s2[i:]
# 5) 比较s1[:i], s2[length_s2-i:] and s1[i:],s2[length_s2:-i]

Update Scramble_String.py