我正在制作一个二叉树,其中包含一个非常大的列表,几乎有 10000 个对象。问题是由于尺寸巨大,我得到了最大递归错误。它特别发生在二叉树类中,其中调用 TreeNode 来创建新对象。我不确定如何在没有递归的情况下实现这一点,因为这似乎是实现代码的最简单方法。
class TreeNode:
def __init__(self,key,val,left=None,right=None,parent=None):
self.key = key
self.payload = val
self.leftChild = left
self.rightChild = right
self.parent = parent
def hasLeftChild(self):
return self.leftChild
def hasRightChild(self):
return self.rightChild
def isLeftChild(self):
return self.parent and self.parent.leftChild == self
def isRightChild(self):
return self.parent and self.parent.rightChild == self
def isRoot(self):
return not self.parent
def isLeaf(self):
return not (self.rightChild or self.leftChild)
def hasAnyChildren(self):
return self.rightChild or self.leftChild
def hasBothChildren(self):
return self.rightChild and self.leftChild
二叉树:
class BinarySearchTree:
def __init__(self):
self.root = None
self.size = 0
def length(self):
return self.size
def __len__(self):
return self.size
def put(self,key,val):
if self.root:
self._put(key,val,self.root)
else:
self.root = TreeNode(key,val)
self.size = self.size + 1
def _put(self,key,val,currentNode):
if key < currentNode.key:
if currentNode.hasLeftChild():
self._put(key,val,currentNode.leftChild)
else:
currentNode.leftChild = TreeNode(key,val,parent=currentNode)
else:
if currentNode.hasRightChild():
self._put(key,val,currentNode.rightChild)
else:
currentNode.rightChild = TreeNode(key,val,parent=currentNode)
def __setitem__(self,k,v):
self.put(k,v)
def get(self,key):
if self.root:
res = self._get(key,self.root)
if res:
return res.payload
else:
return None
else:
return None
def _get(self,key,currentNode):
if not currentNode:
return None
elif currentNode.key == key:
return currentNode
elif key < currentNode.key:
return self._get(key,currentNode.leftChild)
else:
return self._get(key,currentNode.rightChild)
def __getitem__(self,key):
return self.get(key)
def __contains__(self,key):
if self._get(key,self.root):
return True
else:
return False
最佳答案
将递归方法转换为迭代方法相当简单,例如:
def get(self, key):
node = self.root
while node:
if node.key == key:
return node.payload
elif key < node.key:
node = node.leftChild
else:
node = node.rightChild
return None
def put(self, key, val):
if not self.root:
self.root = TreeNode(key, val)
else:
self._put(key, val, self.root)
self.size = self.size + 1
def _put(self, key, val, currentNode):
while True:
if key < currentNode.key:
if currentNode.hasLeftChild():
currentNode = currentNode.leftChild
else:
currentNode.leftChild = TreeNode(key, val, parent=currentNode)
break
else:
if currentNode.hasRightChild():
currentNode = currentNode.rightChild
else:
currentNode.rightChild = TreeNode(key, val, parent=currentNode)
break
这消除了任何递归(限制)并且同样具有可读性。
关于python - 大列表的最大递归,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53038738/