如果不满足循环条件，如何执行一段代码？

Question

Suppose I have a tree and I want to keep going left from the current node. And if there is no left from the current node, I wanna keep going right. otherwise return. To do something like that I'll have to write something like that:

if (node->left != nullptr)
    while (node->left != nullptr)
        node = node->left;
else if (node->right != nullptr)
    while (node->right != nullptr)
        node = node->right;
else return;

As you can see I had to write the condition twice. Can I write something like this:

while (node->left != nullptr)
    node = node->left;
else while (node->right != nullptr)
    node = node->right;
else return;

Also it would be helpful if I can use while with if. Something like this:

while (...) ...
else if (...) ... 
else while (...) ...
else ...

Edit: Just realized that the example of trees is not a good one. But generally can I do something like that or not?

Answer 1

I mean, this will do what you are asking for

if (node->left != nullptr)
    while (node->left != nullptr)
        node = node->left;
else if (node->right != nullptr)
    while (node->right != nullptr)
        node = node->right;
else return;

but maybe is better to check that node is not nullptr (if you're not doing before)

if (node && node->left != nullptr)
    while (node->left != nullptr)
        node = node->left;
else if (node && node->right != nullptr)
    while (node->right != nullptr)
        node = node->right;
else return;

Answer 2

Yes, this is certainly possible in this case. In fact, you can even avoid writing the if and while twice.

Notice that the only difference between the pieces of code, is which Node member is being used (ie are you branching left or right). So, depending on which branch you want to follow, you can take a pointer to that member, like this:

Node* Node::* branch = node->left != nullptr ? &Node::left : &Node::right;

and then unconditionally follow that branch, like this:

while (node->*branch != nullptr)
    node = node->*branch;

In general, when you have two identical pieces of code, you have to decide on a case by case basis which parts of it can be abstracted, and whether the resulting code is improved by it or not.