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.

Understand the problem:

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public class Solution {     public boolean isScramble(String s1, String s2) {         if (s1.length() != s2.length()) {             return false;         }                   if (s1.length() == 1 && s2.length() == 1) {             if (s1.charAt(0) == s2.charAt(0)) {                 return true;             } else {                 return false;             }         }                   char[] array1 = s1.toCharArray();         char[] array2 = s2.toCharArray();         Arrays.sort(array1);         Arrays.sort(array2);         if (!Arrays.toString(array1).equals(Arrays.toString(array2))) {             return false;         }                             for (int i = 1; i < s1.length(); i++) {             String s11 = s1.substring(0, i);             String s12 = s1.substring(i);             String s21 = s2.substring(0, i);             String s22 = s2.substring(i);                           if (isScramble(s11, s21) && isScramble(s12, s22)) {                 return true;             }                           s21 = s2.substring(0, s2.length() - i);             s22 = s2.substring(s2.length() - i);                           if (isScramble(s11, s22) && isScramble(s12, s21)) {                 return true;             }         }                   return false;     } }