告诉转义字符串中十六进制的结尾(Tell where escaped hexadecimal ends in a string [duplicate])

 您正在使用递归进行DFS（深度优先搜索/遍历）。 
 
 其深度首先是因为递归的工作原理与堆栈的工作方式相同 - 在处理下一个节点之前处理当前节点的子节点 - 所以先深入而不是宽度。 
 
 编辑： 
 
 针对您的评论/更新问题：您的代码将逐项按顺序处理，不会有任何并行处理，也不会涉及“魔术”。 使用递归的遍历等价于使用堆栈（LIFO = last in，first out） - 它只是隐含的。 所以你的方法也可以写成如下，它产生相同的遍历顺序： 
 
public void SomeMethod(TreeNode root)
{
    Stack<TreeNode> nodeStack = new Stack<TreeNode>();
    nodeStack.Push(root);

    while (nodeStack.Count > 0)
    {
        TreeNode node = nodeStack.Pop();
        //do something on item
        //need to push children in reverse order, so first child is pushed last
        foreach (TreeNode item in node.Nodes.Reverse())
            nodeStack.Push(item);
    }
}
 
 我希望这可以更清楚地说明发生了什么 - 在处理控制台时将节点写出来或者实际上一步一步地通过调试器逐步浏览可能对您有用。 
 
 （同样，递归方法和使用堆栈的方法都假设没有循环，也不测试 - 所以假设是树而不是任何图。对于后来的DFS引入visited标志来标记节点已经存在看到） 

You are doing a DFS (Depth first search/traversal) right now using recursion.
 
Its depth first because recursion works the same way as a stack would - you process the children of the current node before you process the next node - so you go for depth first instead of breadth.
 
Edit:
 
In response to your comment / updated question: your code will be processed sequentially item by item, there will be no parallel processing, no "magic" involved. The traversal using recursion is equivalent to using a stack (LIFO = last in, first out) - it is just implicit. So your method could also have been written like the following, which produces the same order of traversal:
 
public void SomeMethod(TreeNode root)
{
    Stack<TreeNode> nodeStack = new Stack<TreeNode>();
    nodeStack.Push(root);

    while (nodeStack.Count > 0)
    {
        TreeNode node = nodeStack.Pop();
        //do something on item
        //need to push children in reverse order, so first child is pushed last
        foreach (TreeNode item in node.Nodes.Reverse())
            nodeStack.Push(item);
    }
}
 
I hope this makes it clearer what is going on - it might be useful for you to write out the nodes to the console as they are being processed or actually walk through step by step with a debugger.
 
(Also both the recursive method and the one using a stack assume there is no cycle and don't test for that - so the assumption is this is a tree and not any graph. For the later DFS introduces a visited flag to mark nodes already seen)