Python實現二叉查找樹
二叉查找樹
- 所有 key 小於 V 的都被存儲在 V 的左子樹
- 所有 key 大於 V 的都存儲在 V 的右子樹
BST 的節點
class BSTNode(object):
def __init__(self, key, value, left=None, right=None):
self.key, self.value, self.left, self.right = key, value, left, right
二叉樹查找
如何查找一個指定的節點呢，根據定義我們知道每個內部節點左子樹的 key 都比它小，右子樹的 key 都比它大，所以 對於帶查找的節點 search_key，從根節點開始，如果 search_key 大於當前 key，就去右子樹查找，否則去左子樹查找
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) # 這里值和key一樣的
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小於key則去右子樹找 因為 左子樹<subtree.key<右子樹
subtree.key大於key則去左子樹找 因為 左子樹<subtree.key<右子樹
: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):
"""
查找樹
: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):
"""
查找最小值的樹
:param subtree:
:return:
"""
if subtree is None:
return None
elif subtree.left is None:
# 找到左子樹的頭
return subtree
else:
return self._bst_min_node(subtree.left)
def bst_min(self):
"""
獲取最小樹的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):
"""
查找最大值的樹
:param subtree:
:return:
"""
if subtree is None:
return None
elif subtree.right is None:
# 找到右子樹的頭
return subtree
else:
return self._bst_min_node(subtree.right)
def bst_max(self):
"""
獲取最大樹的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):
"""
二叉查找樹插入
:param subtree:
:param key:
:param value:
:return:
"""
# 插入的節點一定是根節點，包括 root 為空的情況
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):
# 先去查一下看節點是否已存在
node = self._bst_search(self.root, key)
if node is not None:
# 更新已經存在的 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):
"""
刪除並返回根節點
: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:
# 找到了需要刪除的節點
# 要刪除的節點是葉節點 返回 None 把其父親指向它的指針置為 None
if subtree.left is None and subtree.right is None:
return None
# 要刪除的節點有一個孩子
elif subtree.left is None or subtree.right is None:
# 返回它的孩子並讓它的父親指過去
if subtree.left is not None:
return subtree.left
else:
return subtree.right
else:
# 有兩個孩子，尋找后繼節點替換，並從待刪節點的右子樹中刪除后繼節點
# 后繼節點是待刪除節點的右孩子之后的最小節點
# 中(根)序得到的是一個排列好的列表 后繼節點在待刪除節點的后邊
successor_node = self._bst_min_node(subtree.right)
# 用后繼節點替換待刪除節點即可保持二叉查找樹的特性 左<根<右
subtree.key, subtree.value = successor_node.key, successor_node.value
# 從待刪除節點的右子樹中刪除后繼節點，並更新其刪除后繼節點后的右子樹
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)