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(jié)點(diǎn)呢，根據(jù)定義我們知道每個內(nèi)部節(jié)點(diǎn)左子樹的 key 都比它小，右子樹的 key 都比它大，所以 對于帶查找的節(jié)點(diǎn) search_key，從根節(jié)點(diǎn)開始，如果 search_key 大于當(dāng)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, rightclass 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小于key則去右子樹找 因為 左子樹<subtree.key<右子樹 subtree.key大于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(jié)點(diǎn)一定是根節(jié)點(diǎn)，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(jié)點(diǎn)是否已存在 node = self._bst_search(self.root, key) if node is not None: # 更新已經(jīng)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): ''' 刪除并返回根節(jié)點(diǎn) :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(jié)點(diǎn) # 要刪除的節(jié)點(diǎn)是葉節(jié)點(diǎn) 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(jié)點(diǎn)有一個孩子 elif subtree.left is None or subtree.right is None:# 返回它的孩子并讓它的父親指過去if subtree.left is not None: return subtree.leftelse: return subtree.right else:# 有兩個孩子，尋找后繼節(jié)點(diǎn)替換，并從待刪節(jié)點(diǎn)的右子樹中刪除后繼節(jié)點(diǎn)# 后繼節(jié)點(diǎn)是待刪除節(jié)點(diǎn)的右孩子之后的最小節(jié)點(diǎn)# 中(根)序得到的是一個排列好的列表 后繼節(jié)點(diǎn)在待刪除節(jié)點(diǎn)的后邊successor_node = self._bst_min_node(subtree.right)# 用后繼節(jié)點(diǎn)替換待刪除節(jié)點(diǎn)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(jié)點(diǎn)的右子樹中刪除后繼節(jié)點(diǎn)，并更新其刪除后繼節(jié)點(diǎn)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)

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