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) {
            int len1 = s1.length();
            int len2 = s2.length();
            if(len1!=len2) return false;
            char[] c = new char[26];
            for(int i=0;i<len1;i++){
                c[s1.charAt(i)-'a']++;
            }
            for(int i=0;i<len2;i++){
                c[s2.charAt(i)-'a']--;
            }
            for(int i=0;i<26;i++){
                if(c[i]!=0) return false;
            }
            if(len1==len2 &&len1==1) return true;
            for(int i=0;i<len1;i++){
                boolean res= isScramble(s1.substring(0,i),s2.substring(0,i)) && isScramble(s1.substring(i),s2.substring(i));
                        res |=isScramble(s1.substring(0,i),s2.substring(len2-i))&&isScramble(s1.substring(i),s2.substring(0,len2-i));
                if(res) return true;
            }
            return false;
            
        }
    }