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.
提示：
这道题可以用递归方法较为简单地解决，递归的形式可以这样：
// recursive solution, the idea is based on the binary tree struct.
for (int i = ; i < s1.length(); ++i) {
if (isScramble(s1.substr(, i), s2.substr(, i)) && isScramble(s1.substr(i), s2.substr(i))) {
return true;
}
if (isScramble(s1.substr(, i), s2.substr(s2.length() - i)) && isScramble(s1.substr(i), s2.substr(, s2.length() - i))) {
return true;
}
}
结合题目的意思，如果两个单词是scramble的话，那么他们可能会有非叶子节点进行了逆序的操作，因此我们需要同时判断：
- s1的左子节点与s2的左子节点，s1的右子节点与s2的右子节点
- s1的左子节点与s2的右子节点，s1的右子节点与s2的左子节点
另外在进行递归前，需要对字符串是否相等，长度是否相等，是否是变位词进行判断。
代码：
class Solution {
public:
bool isScramble(string s1, string s2) {
// equal ?
if (s1 == s2) {
return true;
}
// if length are differrent, they can not be scramble
if (s1.length() != s2.length()) {
return false;
}
// just like anagram
vector<int> a(, );
for (int i = ; i < s1.length(); ++i) {
++a[s1[i]];
--a[s2[i]];
}
for (int i = ; i < s1.length(); ++i) {
if (a[s1[i]] != ) {
return false;
}
}
// recursive solution, the idea is based on the binary tree struct.
for (int i = ; i < s1.length(); ++i) {
if (isScramble(s1.substr(, i), s2.substr(, i)) && isScramble(s1.substr(i), s2.substr(i))) {
return true;
}
if (isScramble(s1.substr(, i), s2.substr(s2.length() - i)) && isScramble(s1.substr(i), s2.substr(, s2.length() - i))) {
return true;
}
}
return false;
}
};