另类标题：遍历两个节点之间树中的所有节点。
我有一个 DOM 树（有序树，其中每个级别的子元素都有特定的顺序）。我在树中得到了
startstopstartstopstop这是一个众所周知的（命名的？）问题，它具有遍历算法，可能会计算共享祖先或类似的东西来确定是否遍历节点的后代？
示例0（同级）：
<body>
<p>ignore this</p>
<div id="start">start here</div>
<p id="A">visit A</p>
<p id="B">visit B</p>
<p id="C">visit C <span>but <b>not</b> <i>all this</i></span></p>
<div id="stop">stop here</div>
<p>ignore this</p>
</body>
在示例 0 中，我想按顺序访问以下内容：
#A#B#C示例1（止损位更深）：
<body>
<p>ignore this</p>
<div><h2><span id="start">start here</span> A1</h2> A2</div>
<section id="B"><many><more><levels><of><content>…</many></section>
<section id="C">
<div id="D"><many><more><levels><of><content>…</many></div>
<div id="E">
<div id="F"><many><more><levels><of><content>…</many></div>
<div id="stop">stop here</div>
<p>ignore this</p>
</div>
</section>
<p>ignore this</p>
</body>
在示例 1 中，我想按顺序访问以下内容：
A1A2#B#C#D#E#F示例2（止损更高）：
<body>
<p>ignore this</p>
<div>
<h2>
<span id="start">start here</span>
<div id="A">visit A</div>
<section id="B"><many><more><levels><of><content>…</many></section>
</h2>
<section id="C"><many><more><levels><of><content>…</many></section>
</div>
<section id="D"><many><more><levels><of><content>…</many></section>
<div id="stop">stop here</div>
<p>ignore this</p>
</body>
在示例 2 中，我想按顺序访问以下内容：
#A#B#C#D这是一个似乎可以做我想要的事情的算法，用伪代码表示：
results = []
ancestors = get_ancestors_of(stop_node)
current = start_node
while current and current != stop_node:
if current in ancestors:
current = current.children[0]
else:
if current != start_node:
results.add(current)
if current.next_sibling:
current = current.next_sibling
else:
while current and !current.next_sibling:
current = current.parent
if current:
current = current.next_sibling
return results
这是 Ruby 中的方法，它接受一个块并在找到节点时生成节点：
# Traverse nodes between two in the tree, yielding each to the block
# but without traversing the descendants that are wholly between the two.
def traverse_between(start_node, end_node)
# Get all ancestors of the end_node
end_node_ancestors = end_node ? end_node.ancestors.to_set : Set.new
current = start_node
while current && current != end_node
# Traverse children if the end node is in there somewhere.
# We don't yield ancestors of the end_node, since
# they implicitly contain the end_node and possibly more.
if end_node_ancestors.include?(current)
current = current.element_children.first
else
yield current unless current == start_node
if current.next_sibling
current = current.next_sibling
else
# Move up to the next ancestor with a sibling; stop if we reach the root
while current && !current.next_sibling
current = current.parent
return if current.is_a?(Nokogiri::XML::Document)
end
current = current.next_sibling if current
end
end
end
end