原题地址：https://oj.leetcode.com/problems/scramble-string/

题意：

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.

解题思路：二叉树的问题一般使用递归来解决。

代码：

class Solution:
    # @return a boolean
    def isScramble(self, s1, s2):
        if len(s1)!=len(s2): return False
        if s1==s2: return True
        l1=list(s1); l2=list(s2)
        l1.sort();l2.sort()
        if l1!=l2: return False
        length=len(s1)
        for i in range(1,length):
            if self.isScramble(s1[:i],s2[:i]) and self.isScramble(s1[i:],s2[i:]): return True
            if self.isScramble(s1[:i],s2[length-i:]) and self.isScramble(s1[i:],s2[:length-i]): return True
        return False