LeetCode - 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==null || s2==null || s1.length()!=s2.length()) {

            return false;

        }

        char c1[] = s1.toCharArray();

        char c2[] = s2.toCharArray();

        Arrays.sort(c1);

        Arrays.sort(c2);

        String st1 = new String(c1);

        String st2 = new String(c2);

        if(!st1.equals(st2)) {

            return false;

        } else if(st1.length()<=2) {

            return true;

        }

        for(int i=0;i<s1.length()-1;i++) {

            if( (isScramble(s1.substring(0, i+1), s2.substring(0, i+1)) &&

                isScramble(s1.substring(i+1), s2.substring(i+1))) ||

                (isScramble(s1.substring(0, i+1), s2.substring(s1.length()-i-1)) &&

                isScramble(s1.substring(i+1), s2.substring(0, s1.length()-i-1))

                    )

                ) {

                    return true;

                }

        }

        return false;

    }

}

======

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1 == null && s2 == null) {
            return true;
        } else if (s1 == null || s2 == null) {
            return false;
        }
        if (s1.equals(s2)) return true;
        char[] cs1 = s1.toCharArray();
        char[] cs2 = s2.toCharArray();
        Arrays.sort(cs1);
        Arrays.sort(cs2);
        if (!(new String(cs1)).equals(new String(cs2))) {
            return false;
        }
        for (int i = 1; i < s1.length(); i++) {
            if ((isScramble(s1.substring(0, i), s2.substring(0, i)) && isScramble(s1.substring(i), s2.substring(i))) ||
                (isScramble(s1.substring(0, i), s2.substring(s2.length() - i)) && isScramble(s1.substring(i), s2.substring(0, s2.length() - i)))
                ) return true;
        }
        return false;
    }
}