Scramble String

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.

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.equals(s2)) return true;// (s1 == s2) would be time limit exceeded, it compares memory location
        
        int len = s1.length();
        int sum1 = 0;
        int sum2 = 0;
        for(int i = 0; i < len; i++){
            sum1 += s1.charAt(i) - 'a';
            sum2 += s2.charAt(i) - 'a';
        }
        if(sum1 != sum2) return false;
        
        for(int i = 1; i < len; i++){
            if(isScramble(s1.substring(0,i), s2.substring(0,i)) && isScramble(s1.substring(i), s2.substring(i)))
                return true;
            if(isScramble(s1.substring(0,i), s2.substring(len-i)) && isScramble(s1.substring(i), s2.substring(0,len-i)))
                return true;
        }
        return false;
    }
}