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) {       //we have assume that s1 and s2 have the same length and are non-empty				if(s1.equals(s2)) return true;				//check two string contains the same characters		char[] s1char = s1.toCharArray();		char[] s2char = s2.toCharArray();		Arrays.sort(s1char);		Arrays.sort(s2char);		for(int i = 0; i < s1.length(); ++i){			if(s1char[i] != s2char[i])				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)))				return true;			if(isScramble(s1.substring(0, i), s2.substring(s1.length() - i)) && isScramble(s1.substring(i), s2.substring(0,s1.length() - i)))				return true;		}		return false;    }}