leetcode.87. 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.

class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1 == s2)
return true;
if(s1.size() != s2.size())
return false;

vector<int> count(26, 0);
for(int i = 0; i < s1.size(); i ++)
{
count[s1[i]-'a'] ++;
count[s2[i]-'a'] --;
}
for(int i = 0; i < 26; i ++)
{
if(count[i] != 0)
return false;
}

for(int i = 1; i < s1.size(); i ++)
{
if(
(isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))
|| (isScramble(s1.substr(0, i), s2.substr(s1.size()-i)) && isScramble(s1.substr(i), s2.substr(0, s1.size()-i)))
)
return true;
}
return false;

}
};