在python中删除二进制搜索树中的节点？
[英]Deleting nodes in a binary search tree in python?
问题描述
所以我在我的计算机科学课上遇到一个问题，它要求我扩展我们以前编写的二进制搜索树类，并添加一种从树中删除节点的方法，如果该节点有这个必须考虑的话0,1或2个孩子。 在查看条件和内容后，我得出结论，我需要找到指定节点的父节点才能执行此操作，因为已删除节点的子节点将分配给父节点。 这就是我被困住的地方，因为我们刚刚开始使用节点和链表我还是比较新的，这就是为什么我不知道如何跟踪父节点。
这是BST类的代码：
进口副本
class _BSTNode:
def __init__(self, value):
"""
-------------------------------------------------------
Creates a node containing a copy of value.
Use: node = _BSTNode( value )
-------------------------------------------------------
Preconditions:
value - data for the node (?)
Postconditions:
Initializes a BST node containing value. Child pointers are None, height is 1.
-------------------------------------------------------
"""
self._value = copy.deepcopy(value)
self._left = None
self._right = None
self._height = 1
return
def _update_height(self):
"""
-------------------------------------------------------
Updates the height of the current node.
Use: node._update_height()
-------------------------------------------------------
Postconditions:
_height is 1 plus the maximum of the node's (up to) two children.
-------------------------------------------------------
"""
if self._left is None:
left_height = 0
else:
left_height = self._left._height
if self._right is None:
right_height = 0
else:
right_height = self._right._height
self._height = max(left_height, right_height) + 1
return
def __str__(self):
"""
-------------------------------------------------------
DEBUG: Prints node height and value.
-------------------------------------------------------
"""
return "h: {}, v: {}".format(self._height, self._value)
class BST:
def __init__(self):
"""
-------------------------------------------------------
Initializes an empty BST.
Use: bst = BST()
-------------------------------------------------------
Postconditions:
Initializes an empty bst.
-------------------------------------------------------
"""
self._root = None
self._count = 0
return
def is_empty(self):
"""
-------------------------------------------------------
Determines if bst is empty.
Use: b = bst.is_empty()
-------------------------------------------------------
Postconditions:
returns:
True if bst is empty, False otherwise.
-------------------------------------------------------
"""
return self._root is None
def __len__(self):
"""
-------------------------------------------------------
Returns the number of nodes in the BST.
Use: n = len( bst )
-------------------------------------------------------
Postconditions:
returns:
the number of nodes in the BST.
-------------------------------------------------------
"""
return self._count
def insert(self, value):
"""
-------------------------------------------------------
Inserts a copy of value into the bst.
Use: b = bst.insert( value )
-------------------------------------------------------
Preconditions:
value - data to be inserted into the bst (?)
Postconditions:
returns:
inserted - True if value is inserted into the BST,
False otherwise. Values may appear only once in a tree. (bool)
-------------------------------------------------------
"""
self._root, inserted = self._insert_aux(self._root, value)
return inserted
def _insert_aux(self, node, value):
"""
-------------------------------------------------------
Inserts a copy of _value into node.
Private recursive operation called only by insert.
Use: node, inserted = self._insert_aux( node, value )
-------------------------------------------------------
Preconditions:
node - a bst node (_BSTNode)
value - data to be inserted into the node (?)
Postconditions:
returns:
node - the current node (_BSTNode)
inserted - True if value is inserted into the BST,
False otherwise. Values may appear only once in a tree. (boolean)
-------------------------------------------------------
"""
if node is None:
# Base case: add a new node containing the value.
node = _BSTNode(value)
self._count += 1
inserted = True
elif value < node._value:
# General case: check the left subtree.
node._left, inserted = self._insert_aux(node._left, value)
elif value > node._value:
# General case: check the right subtree.
node._right, inserted = self._insert_aux(node._right, value)
else:
# Base case: value is already in the BST.
inserted = False
if inserted:
# Update the current node height if any of its children have been changed.
node._update_height()
return node, inserted
def retrieve(self, key):
"""
-------------------------------------------------------
Retrieves a copy of a value matching key in a BST. (Iterative)
Use: v = bst.retrieve( key )
-------------------------------------------------------
Preconditions:
key - data to search for (?)
Postconditions:
returns:
value - value in the node containing key, otherwise None (?)
-------------------------------------------------------
"""
node = self._root
value = None
while node is not None and value is None:
if key < node._value:
node = node._left
elif key > node._value:
node = node._right
else:
value = copy.deepcopy(node._value)
return value
def inorder(self):
"""
-------------------------------------------------------
Prints the contents of the tree in inorder order.
-------------------------------------------------------
Postconditions:
The contents of the tree are printed inorder.
-------------------------------------------------------
"""
if self._root is not None:
self._inorder_aux(self._root)
return
def _inorder_aux(self, node):
"""
-------------------------------------------------------
Prints the contents of the subtree at node.
-------------------------------------------------------
Preconditions:
node - a bst node (_BSTNode)
Postconditions:
prints:
the values at the subtree of node in inorder
-------------------------------------------------------
"""
if node._left is not None:
self._inorder_aux(node._left)
print(node._value)
if node._right is not None:
self._inorder_aux(node._right)
return
def postorder(self):
"""
-------------------------------------------------------
Prints the contents of the tree in postorder order.
-------------------------------------------------------
Postconditions:
The contents of the tree are printed postorder.
-------------------------------------------------------
"""
if self._root is not None:
self._postorder_aux(self._root)
return
def _postorder_aux(self, node):
"""
-------------------------------------------------------
Prints the contents of the subtree at node.
-------------------------------------------------------
Preconditions:
node - a bst node (_BSTNode)
Postconditions:
prints:
the values at the subtree of node in postorder
-------------------------------------------------------
"""
if node._left is not None:
self._preorder_aux(node._left)
if node._right is not None:
self._preorder_aux(node._right)
print(node._value)
return
def preorder(self):
"""
-------------------------------------------------------
Prints the contents of the tree in preorder order.
-------------------------------------------------------
Postconditions:
The contents of the tree are printed preorder.
-------------------------------------------------------
"""
if self._root is not None:
self._preorder_aux(self._root)
return
def _preorder_aux(self, node):
"""
-------------------------------------------------------
Prints the contents of the subtree at node.
-------------------------------------------------------
Preconditions:
node - a bst node (_BSTNode)
Postconditions:
prints:
the values at the subtree of node in preorder
-------------------------------------------------------
"""
print(node._value)
if node._left is not None:
self._postorder_aux(node._left)
if node._right is not None:
self._postorder_aux(node._right)
return
def traverse(self):
"""
---------------------------------------------------------
Returns the contents of bst in an array in sorted order.
---------------------------------------------------------
Postconditions:
returns:
a - an array containing the contents of bst in sorted order.
---------------------------------------------------------
"""
a = []
self._traverse_aux(self._root, a)
return a
def _traverse_aux(self, node, a):
"""
---------------------------------------------------------
Traverse the node's subtree in inorder, adding the contents of
each node to an array
---------------------------------------------------------
Preconditions:
node - the root of a subtree (_BSTNode)
Postconditions:
a - contains the contents of the subtree of node
in sorted order.
---------------------------------------------------------
"""
if node._left is not None:
self._traverse_aux(node._left, a)
a.append(node._value)
if node._right is not None:
self._traverse_aux(node._right, a)
return
def is_identical(self, rhs):
"""
---------------------------------------------------------
Determines whether two BSTs are identical.
Use: b = bst.is_identical( rhs )
-------------------------------------------------------
Preconditions:
rhs - another bst (BST)
Postconditions:
returns:
identical - True if this bst contains the same values
in the same order as rhs, otherwise returns False.
-------------------------------------------------------
"""
identical = self._is_identical_aux(self._root, rhs._root)
return identical
def _is_identical_aux(self, node1, node2):
"""
---------------------------------------------------------
Determines whether two subtrees are identical.
Use: b = bst.is_identical( node1, node2 )
-------------------------------------------------------
Preconditions:
node1 - node of the current BST (_BSTNode)
node2 - node of the rhs BST (_BSTNode)
Postconditions:
returns:
result - True if this stubtree contains the same values as rhs
subtree in the same order, otherwise returns False.
-------------------------------------------------------
"""
result = True
if node1._value != node2._value:
result = False
if node1._left is not None and node2._left is not None and result == True:
result = self._is_identical_aux(node1._left, node2._left)
if node1.right is not None and node2._right is not None and result == True:
result = self._is_identical_aux(node1._right, node2._right)
return result
def leaf_count(self):
"""
---------------------------------------------------------
Returns the number of leaves (nodes with no children) in bst.
Use: n = bst.leaf_count()
(Recursive algorithm)
---------------------------------------------------------
Postconditions:
returns:
n - number of nodes with no children in bst.
---------------------------------------------------------
"""
n = self._leaf_count_aux(self._root)
return n
def _leaf_count_aux(self, node):
"""
---------------------------------------------------------
Returns the number of leaves (nodes with no children) in bst.
Use: n = bst.leaf_count()
(Recursive algorithm)
---------------------------------------------------------
Preconditions:
node - a BST node (_BSTNode)
Postconditions:
returns:
n - number of nodes with no children below node.
---------------------------------------------------------
"""
count = 0
if node._left is None and node._right is None:
count += 1
if node._left is not None and node._right is None:
count += self._leaf_count_aux(node._left)
if node._left is None and node._right is not None:
count += self._leaf_count_aux(node._right)
if node._left is not None and node._right is not None:
count += self._leaf_count_aux(node._left)
count += self._leaf_count_aux(node._right)
return count
def _delete_node(self, node):
"""
-------------------------------------------------------
Removes a node from the bst.
Use: node = self._delete_node(node)
-------------------------------------------------------
Preconditions:
node - the bst node to be deleted (_BSTNode)
Postconditions:
returns:
node - the node that replaces the deleted node. This node is the node
with the maximum value in the current node's left subtree (_BSTNode)
-------------------------------------------------------
"""
# Your code here.
return node
def parent_i(self, key):
"""
---------------------------------------------------------
Returns the value of the parent node of a key node in a bst.
---------------------------------------------------------
Preconditions:
key - a key value (?)
Postconditions:
returns:
value - a copy of the value in a node that is the parent of the
key node, None if the key is not found.
---------------------------------------------------------
"""
# Your code here
return value
BST类中的最后两个函数parent_i和_delete_node是我正在尝试编写的，注释是我的老师为我们需要编写的函数编写的条件。 我想如果我找到了获取父母的方法，那么删除功能应该很容易，但是现在我不知道从哪里开始。 非常感谢帮助！ 中间的几个函数如min_i和max_i也应该是这样的，可能是我们班级的未来作业或实验室供我们编写。
提前致谢！
编辑：确定所以我现在理解了这个结构，但现在我很困惑，如果我尝试删除一个有2个孩子的节点，并且2个孩子中较高的一个也有2个孩子，依此类推？ 我应该怎样做才能适用于所有尺码的所有款式？
1 个解决方案
解决方案1
0 已采纳 2015-03-05 23:33:23
要确定节点的父节点，请将变量最初设置为
None。 现在，当您对节点进行二进制搜索时，将该变量设置为您之前使用的任何节点，一旦找到该节点，此变量应该为您提供父节点当您尝试删除节点时，如您所述，有3种情况。 只需用1个孩子替换已删除的节点，即可轻松解决前2个问题。 棘手的情况是你有2个孩子。 有两种方法可以解决这个问题，你可以用它的顺序替换那个节点，即它后面的节点（这个节点的右边孩子最左边的孩子），或者用它的前身替换它，即右边的节点在这之前（这个节点的左边孩子的最右边的孩子）。
这只是实际发生的事情的快速浏览，但现在应该很明显，您不需要显式查找节点的父节点以删除该节点，只需跟踪上一个节点前往需要删除的节点就足够了。
有关详细信息，请参阅此示例
Python 3：二进制搜索树，大小函数不返回节点数
[英]Python 3: Binary search tree, size-function not returning number of nodes
在Python中使用顺序遍历（递归）打印二进制搜索树节点
[英]Printing binary search tree nodes using inorder traversal (recursion) in Python
从二进制搜索树中删除时，为什么需要递归返回节点？
[英]Why do I need to recursively return nodes when deleting from a binary search tree?
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