開始的想法：暴力模擬

直接模擬，題目的意思是根節點下的兩個孩子可以交換，就遞迴模擬。然而後面發現根節點的位置似乎並不是固定的，故失敗。
我的程式碼：錯誤程式碼

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;


class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1 == s2)
            return
 
 true;
        return judgeScramble(0,s1.length()-1,s1,s2);
    }

    bool judgeScramble(int from,int to,string str,string pattern)
    {
        cout<<from<<" "<<to<<" "<<str.size()<<endl;
        int mid = (from+to-1)/2;//aim root
        //exchange its two children
        string newstr =
 
 str.substr(0,from)+str.substr(mid+1,to-mid)+str.substr(from,mid+1-from)+str.substr(to+1);
        cout<<newstr<<endl;
        if(newstr == pattern)
            return true;

        bool s1 = false,s2 = false,s3 = false,s4 = false;
        if(mid-from>=1)
        {
            s1 = judgeScramble
 
(from,from+to-mid-1,newstr,pattern);
            s2 = judgeScramble(from,mid,str,pattern);
        }

        if(s1||s2)
            return true;

        if(to-mid>1)
        {
            s3 =  judgeScramble(from+to-mid,to,newstr,pattern);
            s4 = judgeScramble(mid+1,to,str,pattern);
        }


        return s3||s4;
    }
};

int main(void)
{
    Solution s;
    cout<<s.isScramble("abb","bab")<<endl;

    return 0;
}

動態規劃

既然根節點可以交換，那就去反覆列舉根節點（根節點可以從第一個點到最後一個，只要保證孩子的長度大於等於1即可），然後判斷如果發生交換時兩個子串是否符合條件，如果沒有發生交換時兩個子串是否滿足條件。只要有一個滿足條件即可。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1==s2)
            return true;
            
        int len = s1.length();
        int count[26] = {0};
        for(int i=0; i<len; 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<=len-1; i++)
        {
            if( isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i)))
                return true;
            if( isScramble(s1.substr(0,i), s2.substr(len-i)) && isScramble(s1.substr(i), s2.substr(0,len-i)))
                return true;
        }
        return false;
    }
};