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()==0 || s2.size()==0)            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 = 1;n < len;n++){            if(isScramble(s1.substr(0,n),s2.substr(0,n)) && isScramble(s1.substr(n,len-n),s2.substr(n,len-n)))                return true;            if(isScramble(s1.substr(0,n),s2.substr(len-n,n)) && isScramble(s1.substr(n,len-n),s2.substr(0,len-n)))                return true;                }//end for        return false;    }//end func};