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) {

        if(s1.size()== || s2.size()==)

            return false;

        if(s1 == s2)

            return true;

        string a1 = s1,a2 = s2;

        sort(a1.begin(),a1.end());

        sort(a2.begin(),a2.end());

        if(a1!= a2)

            return false;

        int len = s1.size();

        for(int n = ;n < len;n++){

            if(isScramble(s1.substr(,n),s2.substr(,n)) && isScramble(s1.substr(n,len-n),s2.substr(n,len-n)))

                return true;

            if(isScramble(s1.substr(,n),s2.substr(len-n,n)) && isScramble(s1.substr(n,len-n),s2.substr(,len-n)))

                return true;

        }//end for

        return false;

    }//end func

};