c++ - 了解c++代码:使用递归的汉诺塔

我正在学习有关c++中的递归的信息，但是被以下用于解决河内塔问题的c++代码所困扰。

void Hanoi(int m, string start, string middle, string end){
    cout << "m is equal to: " << m << endl;
    if(m == 1){
        cout << "Move Disc " << " from " << start << "  to " << end << endl;
    }
    else{
        Hanoi(m-1,start,end,middle);
        cout << "Move disc " << m << " from " << start << " to " << end << endl;
        Hanoi(m-1,middle,start,end);
    }
}
int main(){
    int discs = 3;
    Hanoi(discs, "start","middle","end");

}

代码的输出如下:

m is equal to: 3
m is equal to: 2
m is equal to: 1
Move Disc  from start  to end
Move disc 2 from start to middle
m is equal to: 1
Move Disc  from end  to middle
Move disc 3 from start to end
m is equal to: 2
m is equal to: 1
Move Disc  from middle  to start
Move disc 2 from middle to end
m is equal to: 1
Move Disc  from start  to end

我的一般问题是我不了解递归的工作方式。为什么在执行“if”语句之前将m设为1？ m如何回到2？

最佳答案

如果将其打印为树，则会得到以下内容:

main
  |--> hanoi(3, ...)
  |      |
  |      |--> hanoi(2, ...)
  |      |     |
  |      |     |--> hanoi(1, ...)
  |      |     |--> hanoi(1, ...)
  |      |<----|
  |      |--> hanoi(2, ...)
  |      |     |
  |      |     |--> hanoi(1, ...)
  |      |     |--> hanoi(1, ...)
  |      |<----|
  |<-----|
  |

对于hanoi(m, ...)的每次调用，除非m == 1，否则它将两次调用hanoi(m-1，...)...在第一个调用中，它将再次调用hanoi(m-1，...)...直到m是1。

因此，当m为2时向后走，它将连续两次调用hanoi(1，...):

   hanoi(2, ...)
      hanoi(1, ...)
      hanoi(1, ...)

当m为3时，它将连续两次调用hanoi(2，...)，因此:

hanoi(3, ...)
   hanoi(2, ...)
      hanoi(1, ...)
      hanoi(1, ...)
   hanoi(2, ...)
      hanoi(1, ...)
      hanoi(1, ...)