二叉查找树，所有左子树key的值都比根小，所有右子树都比他大。
中序遍历，就是从小到大。
BST
查找操作，从根节点开始，search_key大于当前key，去右子树找，否则去左子树找。
插入操作。
class BSTNode(object):
def __init__(self, key, value, left=None, right=None):
self.key, self.value, self.left, self.right = key, value, left, right
class BST(object):
def __init__(self, root=None):
self.root = root
@classmethod
def build_from(cls, node_list):
cls.size = 0
key_to_node_dict = {}
for node_dict in node_list:
key = node_dict['key']
key_to_node_dict[key] = BSTNode(key, value=key)
for node_dict in node_list:
key = node_dict['key']
node = key_to_node_dict[key]
if node_dict["is_root"]:
root = node
node.left = key_to_node_dict.get(node_dict['left'])
node.right = key_to_node_dict.get(node_dict['right'])
cls.size += 1
return cls(root)
"""查找操作"""
def _bst_search(self, subtree, key):
# 辅助函数，不在外部使用
if subtree is None: # 没找到
return None
elif key < subtree.key:
return self._bst_search(subtree.left, key)
elif key > subtree.key:
return self._bst_search(subtree.right, key)
else:
return subtree
def get(self, key, default=None):
node = self._bst_search(self.root, key)
if node is None:
return default
else:
return node.value
"""获取最大和最小key的节点"""
def _bst_min_node(self, subtree):
if subtree is None:
return None
elif subtree.left is None: # 一直往左子树找
return subtree
else:
return self._bst_min_node(subtree.left)
def bst_min(self):
node = self._bst_min_node(self.root)
return node.value if node else None
"""插入操作
需要一直保证左子树小，右子树比根节点大的结构，
观察发现每次插入在叶子节点。
"""
def _bst_insert(self, subtree, key, value):
if subtree is None:
subtree = BSTNode(key, value)
elif key < subtree.key:
subtree.left = self._bst_insert(subtree.left, key, value)
elif key > subtree.key:
subtree.right = self._bst_insert(subtree.right, key, value)
return subtree
def add(self, key, value):
node = self._bst_search(self.root, key)
# 先查找这个节点是否存在，有的话就返回False
if node is not None:
node.value = value
return False
else:
self.root = self._bst_insert(self.root, key, value)
self.size += 1
return True
"""删除节点，较复杂
删除叶节点，（设左或右子树节点为None）
删除只有一个孩子的节点，（把子节点上提）
删除有两个孩子的节点（定义逻辑前任和后继，找到后继进行替换）
找后继：一直查找他右子树的右节点的左子树
"""
def _bst_remove(self, subtree, key):
if subtree is None:
return None
elif key < subtree.key:
subtree.left = self._bst_remove(subtree.left, key)
return subtree
elif key > subtree.key:
subtree.right = self._bst_remove(subtree.right, key)
# return subtree
else: # 找到需要删除的节点
if subtree.left is None and subtree.right is None: # 叶子节点
return None
elif subtree.left is None or subtree.right is None: # 有一个孩子
if subtree.left is not None:
return subtree.left # 返回他的孩子，让他的父亲指过去
else:
return subtree.right
else: # 两个孩子，寻找后继节点并且替换
successor_node = self._bst_min_node(subtree.right)
subtree.key, subtree.value = successor_node.key, successor_node.value
subtree.right = self._bst_remove(subtree.right, successor_node.key)
return subtree
def remove(self, key):
assert key in self
self.size -= 1
return self._bst_remove(self.root, key)
NODE_LIST = [
{'key': 60, 'left': 12, 'right': 90, 'is_root': True},
{'key': 12, 'left': 4, 'right': 41, 'is_root': False},
{'key': 4, 'left': 1, 'right': None, 'is_root': False},
{'key': 1, 'left': None, 'right': None, 'is_root': False},
{'key': 41, 'left': 29, 'right': None, 'is_root': False},
{'key': 29, 'left': 23, 'right': 37, 'is_root': False},
{'key': 23, 'left': None, 'right': None, 'is_root': False},
{'key': 37, 'left': None, 'right': None, 'is_root': False},
{'key': 90, 'left': 71, 'right': 100, 'is_root': False},
{'key': 71, 'left': None, 'right': 84, 'is_root': False},
{'key': 100, 'left': None, 'right': None, 'is_root': False},
{'key': 84, 'left': None, 'right': None, 'is_root': False},
]
def test_bst_tree():
bst = BST.build_from(NODE_LIST)
for node_dict in NODE_LIST:
key = node_dict['key']
assert bst.get(key) == key
assert bst.size == len(NODE_LIST)
#
assert bst.get(-1) is None
assert bst.bst_min() == 1
#
bst.add(0, 0)
assert bst.bst_min() == 0
#
bst.remove(12)
assert bst.get(12) is None
bst.remove(1)
assert bst.get(1) is None
bst.remove(29)
assert bst.get(29) is None
我写这段代码有问题，但是我看不出来，希望读者能提供下思路。
报错self 不是一个可迭代对象。
self = <32二叉查找树.BST object at 0x0000029019EC20D0>, key = 12
def remove(self, key):
> assert key in self
E TypeError: argument of type 'BST' is not iterable
32二叉查找树.py:118: TypeError
=============================================== 1 failed in 0.07s ===============================================
