题目：
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.
思路：
题目中的树形结构也许会成为一个思路误区，使你直接建立树形节点去解题。其实，它仅仅是阐释一个递归方法。
但是题目中依然有个线索，及left1节点中字符的个数设为n1，right1节点中字符的个数设为n2；而另一个节点的left2字符个数为n3，right2字符个数为n4
那么，n1在不等于n2的情况下，不是等于n3就是n4.
因此，在递归的比较长度的left和right的子串的时候，截取的时候需要遵循以下规则：
for(int i=1;i<len;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(len-i)) && isScramble(s1.substring(i),s2.substring(0,len-i)))){
return true;
}
}
而每个isScramble在判断是否相同或者可以互相转化的时候，首先判断长度是否相同；其次比较是否相等，相等直接返回true；
否则，继续判断是否所有字母个数都相同。如果对应字母个数相同，则调用上面的循环进行下一层的比较。
AC代码：
public boolean isScramble(String s1, String s2) {
int len = s1.length();
if(len!=s2.length())
return false;
if(s1.equals(s2))
return true;
int[] bools = new int[27];
for(char a:s1.toCharArray()){
bools[a-'a']++;
}
for(char b:s2.toCharArray()){
bools[b-'a']--;
}
for(Integer aa:bools)
if(aa!=0)
return false;
for(int i=1;i<len;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(len-i)) && isScramble(s1.substring(i),s2.substring(0,len-i)))){
return true;
}
}
return false;
}