[英]Iterative preorder k-ary tree traversal
My python code currently prints out the name of each node in a k-ary tree, from the root to leafs. 我的python代码当前打印出从根到叶子的k进制树中每个节点的名称。 However, I would like the name of branch nodes with children > 1 to print out n times;
但是,我希望子节点> 1的分支节点的名称打印n次。 where n = number of children.
其中n =孩子数。
For the above tree 对于上面的树
My code prints the following 我的代码显示以下内容
start
a
b
d
f
end
c
e
end
c
e
end
However, I want it to print the following 但是,我希望它打印以下内容
start
a
b
d
f
end
b
c
e
end
d
c
e
end
I want nodes b and d to print out twice (in the correct order) since they have two children. 我希望节点b和d打印两次(以正确的顺序),因为它们有两个孩子。
I feel like this is more simple than I am making it out to be. 我觉得这比我想像的要简单。 Do I need to add a list of nodes visited?
我是否需要添加已访问节点的列表? But then I would need to know the number of times visited as well?
但是,我还需要知道访问的次数吗?
One caveat is that only nodes with (n.tag == prefix + 'decision' or n.tag == prefix + 'task') will ever have more than one child. 一个警告是,只有具有(n.tag ==前缀+'决策'或n.tag ==前缀+'任务')的节点才会有多个子节点。 So, I could keep a list of decision/task nodes and the number of times they have been visited.
因此,我可以保留一个决策/任务节点及其访问次数的列表。 If the number of times visited == number of children, pop the node from the list?
如果访问次数==子节点数,请从列表中弹出节点?
I feel like I am over complicating this a lot. 我觉得我已经把这复杂化了很多。
This is a simple example, however my code needs to work for k-ary. 这是一个简单的示例,但是我的代码需要适用于kary。 (I know my example tree is only binary).
(我知道我的示例树只是二进制的)。
My code is below: 我的代码如下:
from itertools import izip
import xml.etree.ElementTree as ET
def main():
prefix = "{http://jbpm.org/4.4/jpdl}"
xml = ET.parse("testWF2.xml")
root = xml.getroot()
i = root.findall('*')
# convert list to dictionary indexed by Element.name
temp = []
for it in i:
name = it.get("name")
if (name):
temp.append(name)
else:
tag = it.tag
temp.append(tag.replace(prefix, '')) # if no name exists use tag (ex. start and end)
b = dict(izip(temp, i)) # create the dictionary with key = name
nodes = []
# add root to the list
nodes.append(b["start"])
while(nodes):
n = nodes.pop()
transitions = n.findall(prefix+"transition")
children = []
# get all of n's children
for t in transitions:
children.append(b[t.get("to")])
for child in children:
nodes.append(child) # add child
if (not n.get("name")):
print ("start")
else:
print(n.get("name"))
# end while loop
main()
If anyone needs to see the testWF2.xml file it is pasted here http://bpaste.net/show/160832/ 如果有人需要查看testWF2.xml文件,则将其粘贴在此处http://bpaste.net/show/160832/
For your special requirement, I changed it so that each iteration works with the parent and the child. 根据您的特殊要求,我对其进行了更改,以便每次迭代都可以与父级和子级一起使用。 The output is based on the parent - in that way the parent automatically is output k-times.
输出基于父级-这样,父级会自动输出k次。 It also needs a special case to output the child when there are no further children.
当没有其他孩子时,还需要特殊情况才能输出孩子。
from itertools import izip
import xml.etree.ElementTree as ET
def main():
prefix = "{http://jbpm.org/4.4/jpdl}"
xml = ET.parse("testWF2.xml")
root = xml.getroot()
i = root.findall('*')
# convert list to dictionary indexed by Element.name
temp = []
for it in i:
name = it.get("name")
if name:
temp.append(name)
else:
tag = it.tag
temp.append(tag.replace(prefix, '')) # if no name exists use tag (ex. start and end)
b = dict(izip(temp, i)) # create the dictionary with key = name
nodes = []
# add root to the list
start_pair = (None, b["start"]) # # # # # using pairs
nodes.append(start_pair)
while(nodes):
parent, n = nodes.pop() # # # # # using pairs
transitions = n.findall(prefix+"transition")
children = []
# get all of n's children
for t in transitions:
child = b[t.get("to")]
children.append(child)
nodes.append((n, child)) # add parent/child pair
# only output the parent (thus outputing k times)
try:
print parent.get("name", "start")
except AttributeError:
pass # ignore the start position
# also output the node if it has no children (terminal node)
if len(children) < 1:
print n.get("name", "start")
# end while loop
main()
start
a
b
d
f
end
d
c
e
end
b
c
e
end
Thats not a tree which you have shown in diagram. 那不是您在图中显示的树。 It is a graph.
它是一个图。 You can use DFS to print out your diagram.
您可以使用DFS打印出图表。 change the start and end nodes for each call to DFS as:
将每次对DFS的调用的起始节点和结束节点更改为:
I am sorry, couldn't put it in a nice table. 对不起,不能把它放在漂亮的桌子上。 I hate stack overflow which makes formatting extremely difficult.
我讨厌堆栈溢出,这使得格式化极其困难。
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