C#•9mo ago

simple algorithm question

FUNCTION CountX(t: Tree; x: INTEGER): INTEGER
BEGIN
    IF t = NIL THEN
        CountX:= 0;
    ELSE IF t^.val = x THEN
        CountX:= 1;
    ELSE
        CountX:= CountX(t^.left; x) + CountX(t^.left; x);
END;

FUNCTION CountX(t: Tree; x: INTEGER): INTEGER
BEGIN
    IF t = NIL THEN
        CountX:= 0;
    ELSE IF t^.val = x THEN
        CountX:= 1;
    ELSE
        CountX:= CountX(t^.left; x) + CountX(t^.left; x);
END;

Quick question: In this function lets say that the "else if" comes true in the first node. wouldnt it just skip the "else" and give out CountX = 1 ? so in order to solve that, my recursive call should not be bound by that "else" and should be done anyway (the code is pascal, but that shouldnt matter with this simple algorithm)

1 Reply

oke•9mo ago

chained if statements run from top to bottom, exiting with the first statement that evaluates to true

int num = 3;

if (num == 1)
{
    Console.WriteLine("It's 1");
}
else if (num == 2)
{
    Console.WriteLine("It's 2");
}
else
{
    Console.WriteLine("It's not 1 or 2");
}

// The output would be "It's 2" because its the first expression to evaluate as True

int num = 3;

if (num == 1)
{
    Console.WriteLine("It's 1");
}
else if (num == 2)
{
    Console.WriteLine("It's 2");
}
else
{
    Console.WriteLine("It's not 1 or 2");
}

// The output would be "It's 2" because its the first expression to evaluate as True

this is it in C#

public static int CountX(TreeNode t, int x)
{
    if (t == null)
    {
        return 0;
    }
    else if (t.val == x)
    {
        return 1;
    }
    else
    {
        return CountX(t.left, x) + CountX(t.right, x);
    }
}

public static int CountX(TreeNode t, int x)
{
    if (t == null)
    {
        return 0;
    }
    else if (t.val == x)
    {
        return 1;
    }
    else
    {
        return CountX(t.left, x) + CountX(t.right, x);
    }
}

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simple algorithm question