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.

     bool isScramble(string s1, string s2) {

         string comp1=s1,comp2=s2;

         sort(comp1.begin(),comp1.end());

         sort(comp2.begin(),comp2.end());

         if(comp1!=comp2)

         return false;

         if(comp1.length()==)

         return true;

         int len=s1.length();

         int i;

         for(i=;i<len;i++)

         {

             string s1_left=s1.substr(,i);

             string s1_right=s1.substr(i);

             string s2_left=s2.substr(,i);

             string s2_right=s2.substr(i);

             if(isScramble(s1_left,s2_left)&&isScramble(s1_right,s2_right))

             return true;

             s2_left=s2.substr(len-i);

             s2_right=s2.substr(,len-i);

             if(isScramble(s1_left,s2_left)&&isScramble(s1_right,s2_right))

             return true;

         }

         return false;

     }