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
};

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