输入树结构是按父/子帐户的层次顺序分隔的财务帐户列表。任何给定的帐户都可以有任意数量的父母/子女。在Python结构中，每个子节点都是一个可以包含任意数量的字典和/或文本值的列表。字典表示指向其他帐户的子项，而文本值表示没有其他后代的子项。下面是一些用JSON格式化的示例输入（测试它，请在Python中将其转换回来）：
[
{
"Assets":[
{
"Bank":[
"Car",
"House"
]
},
{
"Savings":[
"Emergency",
{
"Goals":[
"Roof"
]
}
]
},
"Reserved"
]
}
]
在幕后有一个输入文件，其中包含如下所示的帐户定义：
Assets:Bank:House
Assets:Savings:Emergency
Assets:Savigs:Goals:Roof
我有现有的代码来解析和创建上面看到的树结构。
目标：最终目标是通过搜索树来利用给定的字符串输入提供自动完成功能。使用上面的示例输入，以下输入将产生各自的输出：
"Assets" => ["Bank, "Savings", "Reserved"]
"Assets:Bank" => ["Car", "House"]
"Assets:Savings:Goals" => ["Roof"]
部分解决方案：递归是我被绊倒的地方。我能够创建可以处理“root”帐户的结果的代码，但我不确定如何使其递归以提供子帐户的结果。这是代码：
def search_tree(account, tree):
# Check to see if we're looking for a root level account
if isinstance(account, str) and ":" not in account:
# Collect all keys in the child dictionaries
keys = {}
for item in tree:
if isinstance(item, dict):
keys[item.keys()[0]] = item
# Check to see if the input matches any children
if account in keys:
# Collect all children of this account
children = []
for child in keys[account][account]:
if isinstance(child, str):
children.append(child)
else:
children.append(child.keys()[0])
return children
# tree = .....
account = "Assets"
print search_tree(account, tree) # Would produce ["Bank", "Savings", "Reserved"]
# In the future I would provide "Assets:Bank" as the account string and get back the following: ["Car", "House"]
如何将此递归搜索到 n 的孩子？
答案 0 :(得分：2)
我不会真正回答您的问题（关于您的特定标准输出要求），但我将帮助您了解如何搜索树结构
首先描述你的树形结构
现在我们可以开始编写递归解决方案了
def search(key,tree):
if isinstance(tree,(list,tuple)): # this is a tree
for subItem in tree: # search each "node" for our item
result = search(key,subItem)
if result:
return result
elif isinstance(tree,dict): # this is really a node (nodeType1)
nodeName,subTree = next(tree.iteritems())
if nodeName == key: # match ... in your case the key has many parts .. .you just need the "first part"
print "Found:",key
return subTree
else: # did not find our key so search our subtree
return search(key,subTree)
elif isinstance(tree,basestring): #leaf node
if tree == key: # found our key leaf node
print "Found",key
return tree
这实际上只是一个非常通用的解决方案，它可用于搜索单个条目（即“House”或“Accounts”...它不记录用于获得解决方案的路径）< / p>
现在让我们回到检查您的问题陈述
密钥是多部分密钥Part1:part2:part3
所以让我们开始解决这个问题
def search_multipartkey(key,T,separator=":"):
result = T
for part in key.split(separator):
result = search(part,result)
if not result:
print "Unable to find part:",part
return False
else:
print "Found part %s => %s"%(part,result)
return result
你几乎可以肯定地改进了这一点，但这给了一个很好的起点（尽管它可能不像人们所希望的那样递归）
答案 1 :(得分：0)
不完整（超时，但我相信你会设法整合你的测试）：
tree = [
{"Assets": [
{"Bank": [
"Car",
"House"
]
},
{"Savings": [
"Emergency",
{"Goals":
["Roof"]
}
]
},
"Reserved"
]
}
]
def search_tree(account, tree, level):
""" """
print("account", account)
print("tree", tree)
print("level", level)
print("-------------")
if account == []:
return
r = None
for d in tree:
print("a:",account[0])
print("d:",d)
try:
newtree = d[account[0]]
newaccount = account[1:]
print("new:", newtree, newtree )
r = search_tree(newaccount, newtree, level+1)
except Exception as e:
print("failed because:", e)
return r
account = "Assets:Bank"
search_tree(account.split(":"), tree, 0)
输出：
> py -3 t.py
account ['Assets', 'Bank']
tree [{'Assets': [{'Bank': ['Car', 'House']}, {'Savings': ['Emergency', {'Goals': ['Roof']}]}, 'Reserved']}]
level 0
-------------
a: Assets
d: {'Assets': [{'Bank': ['Car', 'House']}, {'Savings': ['Emergency', {'Goals': ['Roof']}]}, 'Reserved']}
new: [{'Bank': ['Car', 'House']}, {'Savings': ['Emergency', {'Goals': ['Roof']}]}, 'Reserved'] [{'Bank': ['Car', 'House']}, {'Savings': ['Emergency', {'Goals': ['Roof']}]}, 'Reserved']
account ['Bank']
tree [{'Bank': ['Car', 'House']}, {'Savings': ['Emergency', {'Goals': ['Roof']}]}, 'Reserved']
level 1
-------------
a: Bank
d: {'Bank': ['Car', 'House']}
new: ['Car', 'House'] ['Car', 'House']
account []
tree ['Car', 'House']
level 2
-------------
a: Bank
d: {'Savings': ['Emergency', {'Goals': ['Roof']}]}
failed because: 'Bank'
a: Bank
d: Reserved
failed because: string indices must be integers
仍然没有测试，但返回你想要的（对于这个案例）：
def search_tree(account, tree, level):
""" """
#print()
#print()
#print("account", account)
#print("tree", tree)
#print("level", level)
#print("-------------")
if account == []:
#print("reached end")
#print("tree", tree)
return tree
r = None
for d in tree:
#print("a:",account[0])
#print("d:",d)
try:
newtree = d[account[0]]
newaccount = account[1:]
#print("new:", newtree, newtree )
r = search_tree(newaccount, newtree, level+1)
except Exception as e:
#print("failed because:", e)
pass
return r
account = "Assets:Bank"
print( search_tree(account.split(":"), tree, 0) ) # --> ['Car', 'House']