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, 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) # 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)

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