我几乎完成了一个项目,该项目使我们创建了一个使用二叉树结构的字典类。然而,我坚持如何实现一个方法,打印出树中的所有元素,我只是没有太多的二叉树经验,所以它如何编码它相当混乱。遍历Python中的二叉树
我想弄明白的方法是一个键方法,它将遍历整个树并返回所有键的列表。我认识的人暗示我应该创建一个递归遍历树并跟踪所有键的私有助手函数。我想创建他正在谈论的内容,但我没有现成的想法来编写代码。任何人都可以帮我编码吗?搞清楚这一点对我来说几乎完成了。
这是我的代码到目前为止。 [Key:Value]
对是元组。我已经编写,并也有从课本上的例子一些帮助构建你在这里看到:
class DictWithTree:
def __init__(self):
self._element = None
self._left = None
self._right = None
self._size = 0
def isempty(self):
if self._element == None:
return True
return False
def __len__(self):
return self._size
def __contains__(self,key):
path = self._tracePath(key)
return path[-1]._size > 0
def _tracePath(self,key): # taken from the textbook example and modified
if len(self) == 0 or key == self._element[0]:
return [self]
elif len(key) < len(self._element[0]):
return [self] + self._left._tracePath(key)
else:
return [self] + self._right._tracePath(key)
def __getitem__(self,key):
if len(self) == 0:
raise KeyError(key)
elif key == self._element[0]:
return self._element[1]
elif key < self._element[0]:
return self._left[key]
elif key > self._element[0]:
return self._right[key]
else:
raise KeyError(key)
def __setitem__(self,key,value):
path = self._tracePath(key)
endOfPath = path[-1]
if endOfPath._element != None:
if endOfPath._element[0] == key:
endOfPath._element = key,value
if endOfPath._size == 0: # a new element
for location in path:
location._size += 1
endOfPath._element = key,value
endOfPath._left = DictWithTree()
endOfPath._right = DictWithTree()
def clear(self):
self._element = None
self._left = None
self._right = None
self._size = 0
def pop(self,key):
value = self[key]
self._remove(key)
return value
def popitem(self): # returns the 'last' item in the dictionary,
if self.isempty(): # (i.e. the largest key in the dictionary)
return KeyError("There are no keys in the dictionary")
elif self._right._element == None:
return self._element
else:
return self._right.popitem()
def _remove(self,key):
path = self._tracePath(key)
endOfPath = path[-1]
if endOfPath._size > 0:
for location in path:
location._size -= 1
if len(endOfPath._left) == 0:
endOfPath._promoteChild(endOfPath._right)
elif len(endOfPath._right) == 0:
endOfPath._promoteChild(endOfPath._left)
else:
endOfPath._element = endOfPath._left.pop()
def _promoteChild(self,child):
self._element = child._element
self._left = child._left
self._right = child._right
为什么你使用自己写的树而不是Python字典? – Will 2011-04-16 18:54:51
这是一个项目,这是我们应该做的。创建一个使用二叉树的字典。 – Eric 2011-04-16 19:02:04
我相信他说这是他的项目规格。 – ninjagecko 2011-04-16 19:02:54