题目：

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

代码：

方法一（用了排序算法的递归)：

class Solution {
public:
    bool isScramble(string s1, string s2) {
         if (s1.size() != s2.size()) return false;
        if (s1 == s2) return true;
        string str1 = s1, str2 = s2;
        sort(str1.begin(), str1.end());
        sort(str2.begin(), str2.end());
        if (str1 != str2) return false;
        for (int i = 1; i < s1.size(); ++i) {
            string s11 = s1.substr(0, i);
            string s12 = s1.substr(i);
            string s21 = s2.substr(0, i);
            string s22 = s2.substr(i);
            if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
            s21 = s2.substr(s1.size() - i);
            s22 = s2.substr(0, s1.size() - i);
            if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
        }
        return false;
    }
};

方法二(没用排序算法的递归):

class Solution {
public:
    bool isScramble(string s1, string s2) {
          if (s1 == s2) return true;
        if (s1.size() != s2.size()) return false;
        int n = s1.size(), m[26] = {0};
        for (int i = 0; i < n; ++i) {
            ++m[s1[i] - 'a'];
            --m[s2[i] - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            if (m[i] != 0) return false;
        }
        for (int i = 1; i < n; ++i) {
            if ((isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i))) || (isScramble(s1.substr(0, i), s2.substr(n - i)) && isScramble(s1.substr(i), s2.substr(0, n - i)))) {
                return true;
            }
        }
        return false;
    }
};