Python实现二叉查找树
二叉查找树
- 所有 key 小于 V 的都被存储在 V 的左子树
- 所有 key 大于 V 的都存储在 V 的右子树
BST 的节点
class BSTNode(object):
def __init__(self, key, value, left=None, right=None):
self.key, self.value, self.left, self.right = key, value, left, right
二叉树查找
如何查找一个指定的节点呢，根据定义我们知道每个内部节点左子树的 key 都比它小，右子树的 key 都比它大，所以 对于带查找的节点 search_key，从根节点开始，如果 search_key 大于当前 key，就去右子树查找，否则去左子树查找
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},
]
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) # 这里值和key一样的
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)
def _bst_search(self, subtree, key):
"""
subtree.key小于key则去右子树找 因为 左子树<subtree.key<右子树
subtree.key大于key则去左子树找 因为 左子树<subtree.key<右子树
:param subtree:
:param key:
:return:
"""
if subtree is None:
return None
elif subtree.key < key:
self._bst_search(subtree.right, key)
elif subtree.key > key:
self._bst_search(subtree.left, key)
else:
return subtree
def get(self, key, default=None):
"""
查找树
:param key:
:param default:
:return:
"""
node = self._bst_search(self.root, key)
if node is None:
return default
else:
return node.value
def _bst_min_node(self, subtree):
"""
查找最小值的树
:param subtree:
:return:
"""
if subtree is None:
return None
elif subtree.left is None:
# 找到左子树的头
return subtree
else:
return self._bst_min_node(subtree.left)
def bst_min(self):
"""
获取最小树的value
:return:
"""
node = self._bst_min_node(self.root)
if node is None:
return None
else:
return node.value
def _bst_max_node(self, subtree):
"""
查找最大值的树
:param subtree:
:return:
"""
if subtree is None:
return None
elif subtree.right is None:
# 找到右子树的头
return subtree
else:
return self._bst_min_node(subtree.right)
def bst_max(self):
"""
获取最大树的value
:return:
"""
node = self._bst_max_node(self.root)
if node is None:
return None
else:
return node.value
def _bst_insert(self, subtree, key, value):
"""
二叉查找树插入
:param subtree:
:param key:
:param value:
:return:
"""
# 插入的节点一定是根节点，包括 root 为空的情况
if subtree is None:
subtree = BSTNode(key, value)
elif subtree.key > key:
subtree.left = self._bst_insert(subtree.left, key, value)
elif subtree.key < key:
subtree.right = self._bst_insert(subtree.right, key, value)
return subtree
def add(self, key, value):
# 先去查一下看节点是否已存在
node = self._bst_search(self.root, key)
if node is not None:
# 更新已经存在的 key
node.value = value
return False
else:
self.root = self._bst_insert(self.root, key, value)
self.size += 1
def _bst_remove(self, subtree, key):
"""
删除并返回根节点
:param subtree:
:param key:
:return:
"""
if subtree is None:
return None
elif subtree.key > key:
subtree.right = self._bst_remove(subtree.right, key)
return subtree
elif subtree.key < key:
subtree.left = self._bst_remove(subtree.left, key)
return subtree
else:
# 找到了需要删除的节点
# 要删除的节点是叶节点 返回 None 把其父亲指向它的指针置为 None
if subtree.left is None and subtree.right is None:
return None
# 要删除的节点有一个孩子
elif subtree.left is None or subtree.right is None:
# 返回它的孩子并让它的父亲指过去
if subtree.left is not None:
return subtree.left
else:
return subtree.right
else:
# 有两个孩子，寻找后继节点替换，并从待删节点的右子树中删除后继节点
# 后继节点是待删除节点的右孩子之后的最小节点
# 中(根)序得到的是一个排列好的列表 后继节点在待删除节点的后边
successor_node = self._bst_min_node(subtree.right)
# 用后继节点替换待删除节点即可保持二叉查找树的特性 左<根<右
subtree.key, subtree.value = successor_node.key, successor_node.value
# 从待删除节点的右子树中删除后继节点，并更新其删除后继节点后的右子树
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)