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:**

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; } }

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