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.

[Thoughts]: Top-Down,  Build approach from subproblems to the final destination。
isScramble(s1, s2) = true,
那么一定有一个位置i, 把s1分成两段，p1 和 p2,  把s2分成两段， q1 和 q2
且要么isScramble(p1, q1) && isScramble(p2, q2)
要么isScramble(p1, q2) && isScramble(p2, q1) // 记得计算substring使p2.len = q1.len.

public class Solution {
    public boolean isScramble(String s1, String s2) {
            if(s1.length()!=s2.length()) return false;
        s1.toLowerCase(); s2.toLowerCase();
        int[] value = new int[26];
        //check if s1 and s2 have same characters. 
        for(int i=0; i<s1.length(); i++){
            value[s1.charAt(i)-'a']++;
            value[s2.charAt(i)-'a']--;
        }
        for(int i=0; i<value.length; i++){
            if(value[i]!=0)
                return false;
        }
        //收敛条件，只有一个char且相等时，返回ture
        if(s1.length()==1)
            return true;
        
        for(int i=0; i<s1.length(); 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(s2.length()-i)) && isScramble(s1.substring(i), s2.substring(0,s2.length()-i)))
                return true;
        }
        
        return false;
    }
}