可视化二叉树的递归

Question

Write a recursive function that displays all the binary (base 2) numbers represented by a string of xs, 0s, and 1s. The xs represent digits that can be either 0 or 1. For example, the string xx represents the numbers 00,01,10,11.

The code works, but I just have a hard time visualizing the intermediate steps. Could someone help me walkthrough?

void get_first_x(char *line,char *line2,char *line3);
void display(char *line);
int main(int argc, const char * argv[]) {

    char line[256];
    printf("Binary number: ");
    scanf("%s",line);
    display(line);
    return 0;
}


void display(char *line){


    char line2[80];
    char line3[80];

    if(strchr(line,'x') == NULL)
    {
        printf("%s\n",line);
        return;
    }

   get_first_x(line,line2,line3);


    display(line2);

    display(line3);

  }

   void get_first_x(char* line,char* line2,char *line3) {

    char* check;

    check = strchr(line,'x');
    *check = '0';

    strcpy(line2,line);

    *check = '1';
    strcpy(line3,line);

}//replacement of x with 0 and 1. One argument produces 2 strings

Here's my take

1st call     display(xx)
2nd call       display(0x)
3rd call          display(00) { print statement/ return}
                     display(1x)
                        display(01) { print statement/return}
                          display(10) { print statement/return}
                           recursion exits
 Input: xx
 output: 00,01,10,11
 I'm not understanding something...here

Answer 1

What you implemented is this (in pseudo code):

display(line) {
  if no_x_in(line) {
     print(line)  // instance output and recursion stop 
  } 
  display(replace_first_x_with_0(line))  // recursive call
  display(replace_first_x_with_1(line))  // recursive call
}

• If the string in line contains no x symbols anymore you can output the string and your recursive descent can stop.

• If not, the problem instance is reduced from a line with n times many x symbols into two smaller instances, each with n - 1 many x symbols,

  • one with the x replaced by a 0 symbol and
  • one with the x replaced by a 1 symbol.

which result into a recursive call each. As there are only finite many x symbols in the finite input string, the recursive calls will stop at some point, and the resulting call tree is finite as well.

For your example the call tree is like this:

display('xx') -> issues calls to display('0x') and display('1x')
|
+-> display('0x') -> issues calls to display('00') and display('01')
|   |
|   +-> display('00') -> output, stop
|   +-> display('01') -> output, stop
|
+-> display('1x') -> issues calls to display('10') and display('11')
    |
    +-> display('10') -> output, stop
    +-> display('11') -> output, stop