Scramble String

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 {
public:
    bool isScramble(string s1, string s2) {
        int len = s1.length();
        int lenS2 = s2.length();
        if (len != lenS2)
        {
            return false;
        }
        if (s1 == s2)
        {
            return true;
        }

        int count[256] = {0};
        for (int i = 0; i < len; i++)
        {
            count[s1[i]]++;
            count[s2[i]]--;
        }

        for (int i = 0; i < 256; i++)
        {
            if (count[i] != 0)
            {
                return false;
            }
        }

        for (int i = 1; i < len; i++)
        {
            if (isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))
            {
                return true;
            }
            if (isScramble(s1.substr(0, i), s2.substr(len-i)) && isScramble(s1.substr(i), s2.substr(0, len-i)))
            {
                return true;
            }
        }

        return false;
    }
};