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; }

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