[Leetcode] scramble string (tree problems are the easiest to use recursion)

Given a stringS1, We may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representationS1="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"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"great".

Given two stringsS1AndS2Of the same length, determine ifS2Is a scrambled stringS1.

class Solution {public:    bool isScramble(string s1, string s2) {        if(s1.size()==0 || s2.size()==0)            return false;        if(s1 == s2)            return true;        string a1 = s1,a2 = s2;        sort(a1.begin(),a1.end());        sort(a2.begin(),a2.end());        if(a1!= a2)            return false;        int len = s1.size();        for(int n = 1;n < len;n++){            if(isScramble(s1.substr(0,n),s2.substr(0,n)) && isScramble(s1.substr(n,len-n),s2.substr(n,len-n)))                return true;            if(isScramble(s1.substr(0,n),s2.substr(len-n,n)) && isScramble(s1.substr(n,len-n),s2.substr(0,len-n)))                return true;                }//end for        return false;    }//end func};