Need help understanding recursion

Question

I am trying to understand how recursion works. I have a basic idea, but the details remain unclear. Here is a simple javascript example:

function sumTo(n){
    if (n > 1){
        return n + sumTo(n-1)
    } else {
        return n
    }
}

sumTo(3);

It is supposed to count all numbers in 3, and the result is 6 (1 + 2 + 3 = 6), but I do not understand how this works.

OK, we start with the if condition. 3> 1, so we return n and call the function again, but what happens inside the brackets?

It looks like this:

3 + sumTo (2) // 3 - 1 = 2

Or we do nothing with n and wait for the following function:

3 + sumTo (n - 1) // n will come later

I was told that the last function will return 1 to the top, but I don’t understand what it will do with this 1.

If there are several step-by-step explanations for the final layouts, please share.

I know that there are many similar questions, but I did not find any help.

UPD: , , , , . , , . , , - , .

, . http://www.integralist.co.uk/posts/js-recursion.html (1, 10) (2, 3).

function sum(x, y) {
    if (y > 0) {
      return sum(x + 1, y - 1);
    } else {
      return x;
    }
}

sum(2, 3);

, if > 0. Y 3, . (x + 1, y - 1), i. . (2 + 1, 3 - 1), .. e sum (3, 2).

(3, 2). , y > 0. Y 2, . (x + 1, y - 1), i. . (3 + 1, 2 - 1), .. e sum (4, 1).

(4, 1). y > 0. Y 1, . (x + 1, y - 1), i. . (4 + 1, 1 - 1), .. e sum (5, 0).

(5, 0). y > 0 . if-else , x, 5. , (2, 3) 5.

sumTo();

function sumTo(n){
    if (n > 1){
        return n + sumTo(n-1)
    } else {
        return n
    }
}

sumTo(3);

sumTo (3). n > 1: 3 > 1 , n + sumTo (n - 1), i. . 3 + sumTo (3 - 1), .. . 3 + sumTo (2).

, sumTo (2).

n > 1: 2 > 1 , n + sumTo (n - 1), i. . 2 + sumTo (2 - 1), i. . 2 + sumTo (1).

, sumTo (1).

, n > 1. 1 > 1 , , if-else, n, i. . 1. , sumTo (1) 1.

sumTo (1) sumTo (2), , , To (1) .

SumTo (2) n + sumTo (n-1), i. . 2 + sumTo (2 - 1), i. . 2 + sumTo (1), i. . 2 + 1. , sumTo (2) 3.

sumTo (2) sumTo (3), , , To (2) .

SumTo (3) n + sumTo (n-1), i. . 3 + sumTo (3 - 1), .. . 3 + sumTo (2), i. . 3 + 3. , sumTo (3) 6. , sumTo (3) 6.

, 3 n, n 1.

, .

Answer 1

:

sumTo(4) = (4 + 3) + (2 + 1) = 10 // 4 + sumTo(3). function called four times
sumTo(3) = (3 + 2) + 1       = 6  // 3 + sumTo(2). called three times
sumTo(2) = (2 + 1)           = 3  // 2 + sumTo(1). called twice
sumTo(1) = (1)               = 1  // called once

, , , , . :

sumTo(1) = 1 + sumTo(0) = 1
sumTo(2) = 2 + sumTo(1) = 3
sumTo(3) = 3 + sumTo(2) = 6
sumTo(4) = 4 + sumTo(3) = 10

, , , .

:

sumTo(3) 3 sumTo(2), sumTo(1) 3, 6.

? , - . , . : ?

. , , , . .

:

function multiplyBy10(i) {
    if ( !i ) return 0;
    return 10+multiplyBy10(i-1);
}

10 . , , , . , , .:)

Answer 2

, sumTo(n) , sumTo(n-1) .
, sumTo(3) sumTo(2),
sumTo(2) sumTo(1),
sumTo(1) 1,
sumTo(2) 2 + 1
sumTo(3) 3 + 2 + 1

Answer 3

You can understand how

sumTo(3);
returns => 3 +  sumTo(2) // n is greater than 1
                returns => 2 +  sumTo(1)
                                returns => 1 // 1 as n is not greater than 1