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.

 

 1     bool isScramble(string s1, string s2) {
 2         string comp1=s1,comp2=s2;
 3         sort(comp1.begin(),comp1.end());
 4         sort(comp2.begin(),comp2.end());
 5         if(comp1!=comp2)
 6         return false;
 7         if(comp1.length()==1)
 8         return true;
 9         int len=s1.length();
10         int i;
11         for(i=1;i<len;i++)
12         {
13             string s1_left=s1.substr(0,i);
14             string s1_right=s1.substr(i);
15             string s2_left=s2.substr(0,i);
16             string s2_right=s2.substr(i);
17             if(isScramble(s1_left,s2_left)&&isScramble(s1_right,s2_right))
18             return true;
19             s2_left=s2.substr(len-i);
20             s2_right=s2.substr(0,len-i);
21             if(isScramble(s1_left,s2_left)&&isScramble(s1_right,s2_right))
22             return true;
23         }
24         return false;
25         
26         
27     }