在递归斐波那契程序中，为什么左子树在右子树之前被调用（类似于后顺序遍历）？

Question

package fibonacci;

public class Fib {

    static int fib(int n)
    {
        System.out.println("fib(" + n + ") called");
        if(n<=1)
        {
            return n;
        }
        int temp =  fib(n-1) + fib(n-2);
        System.out.println("returning to fib(" + n +")" );
        return temp;
    }
    public static void main(String[] args)
    {
        System.out.println("fib(5): " + fib(5));
    }
}

---

Output:

fib(5) called
fib(4) called
fib(3) called
fib(2) called
fib(1) called
fib(0) called
returning to fib(2)
fib(1) called
returning to fib(3)
fib(2) called
fib(1) called
fib(0) called
returning to fib(2)
returning to fib(4)
fib(3) called
fib(2) called
fib(1) called
fib(0) called
returning to fib(2)
fib(1) called
returning to fib(3)
returning to fib(5)
fib(5): 5

PS: Why fib(n-1) is called before fib(n-2)?

Answer 1

This can be explained by the following section of JLS -

The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.

In your case, the left operand is the call to fib(n-1) , so it will be fully evaluated to compute final value, and only then the right operation will be evaluated.

Answer 2

Because the first part that is executed, is called first, for example if you so this :

void dfs(int x) {
   dfs(x-1);
   dfs(x-2);
}

Then dfs(x-1) will called first, so dfs will be called again, and dfs(x-1) will called one more time because it's the first instruction and so on until last dfs(x-1) is called then the program continue with dfs(x-2).

I hope it helps.