使这个函数不递归？

Question

我在这里有一个递归的函数,但我宁愿让它非递归.我只是不确定如何.

  void AguiWidgetManager::recursiveRender(const AguiWidget *root)
{
    //recursively calls itself to render widgets from back to front
    AguiWidget* nonConstRoot = (AguiWidget*)root;
    if(!nonConstRoot->isVisable())
    {
        return;
    }

        clip(nonConstRoot);

        nonConstRoot->paint(AguiPaintEventArgs(true,graphicsContext));

        for(std::vector<AguiWidget*>::const_iterator it = 
            root->getPrivateChildBeginIterator();
            it != root->getPrivateChildEndIterator(); ++it)
        {
            recursiveRender(*it);
        }
        for(std::vector<AguiWidget*>::const_iterator it = 
            root->getChildBeginIterator();
            it != root->getChildEndIterator(); ++it)
        {
            recursiveRender(*it);
        }

}

如果解决方案不适用于迭代器,那也没关系.

谢谢

Answer 1

简单的方法就是维护自己的堆栈.Psuedoish示例代码:

stack s;
while(!s.empty())
{
    root = s.pop();

    //your code here
    //replace each recursive call with s.push(it)
}

另外,抛弃常量是一个糟糕的坏主意.如果你想修改它,它首先不应该是一个const参数.