LeetCode 87: 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 string s1 and s2 are scramble strings, there must be a point that breaks s1 to two parts s11, s12,
        // and a point that breaks s2 to two parts, s21, s22, and isScramble(s11, s21) && isScramble(s12, s22) is true,
        // or isScramble(s11, s22) && isScramble(s12, s21) is true.

        // So we can make it recursively.
        // We just break s1 at different position to check if there exists one position satisfies the requirement.

        // Some checks are needed otherwise it will time out.
        // For example, if the lengths of two strings are different, they can’t be scramble.
        // And if the characters in two strings are different, they can’t be scramble either.
        
        // Step 1. Check lengths.
        if (s1.length() != s2.length())
            return false;
        
        // Step 2. Check equality.    
        if (s1.equals(s2))
            return true;
            
        int L = s1.length();
        
        // Step 3. Check if with the same characters
        int[] c = new int[256];
        
        for (int i = 0; i < L; i++)
        {
            c[s1.charAt(i)]++;
            c[s2.charAt(i)]--;
        }
        
        for (int i = 0; i < 256; i++)
            if (c[i] != 0)
                return false;
            
        // Step 4. Check every dividing point
        for (int i = 1; i < L; i++)
        {
            String s11 = s1.substring(0, i);
            String s12 = s1.substring(i, L);
            String s21 = s2.substring(0, i);
            String s22 = s2.substring(i, L);
            
            if (isScramble(s11, s21) && isScramble(s12, s22))
                return true;
                
            s21 = s2.substring(0, L - i);
            s22 = s2.substring(L - i, L);
            
            if (isScramble(s11, s22) && isScramble(s12, s21))
                return true;
        }
        
        return false;        
    }
}