LeetCode-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 isS(string& s1,string& s2,int begin1,int begin2,int end1,int end2)
    {
        int length1=end1-begin1+1;
        int length2=end2-begin2+1;
        
       
        if(length1!=length2)return false;
        
        if(length1==1){
            if(s1[begin1]==s2[begin2])return true;
            else return false;
        }
        else{
            vector<int> count;
            count.resize(26);
            for(int i=0;i<26;i++)count[i]=0;
            for(int i=begin1;i<=end1;i++){
                count[s1[i]-'a']++;
            }
            for(int i=begin2;i<=end2;i++){
                count[s2[i]-'a']--;
            }
            for(int i=0;i<26;i++){
                if(count[i]!=0){
                    return false;
                }
            }
            for(int i=1;i<length1;i++){
                bool b1,b2,b3,b4;
                int j=length1-i;
                b1=isS(s1,s2,begin1,begin2,begin1+i-1,begin2+i-1);
                b2=isS(s1,s2,begin1+i,begin2+i,end1,end2);
                b3= isS(s1,s2,begin1,end2-i+1,begin1+i-1,end2);
                b4=isS(s1,s2,begin1+i,begin2,end1,begin2+j-1);
                if(
                    b1&&b2
                ||
                    b3&&b4
                )
                return true;
            }
            return false;
        }
    }
    bool isScramble(string s1, string s2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return isS(s1,s2,0,0,s1.length()-1,s2.length()-1);
    }
};