""" File: binary_search_tree.py Created Time: 2022-12-20 Author: a16su (lpluls001@gmail.com) """ import sys, os.path as osp sys.path.append(osp.dirname(osp.dirname(osp.abspath(__file__)))) from modules import * class BinarySearchTree: """二叉搜索树""" def __init__(self, nums: list[int]) -> None: """构造方法""" nums.sort() self.__root = self.build_tree(nums, 0, len(nums) - 1) def build_tree( self, nums: list[int], start_index: int, end_index: int ) -> TreeNode | None: """构建二叉搜索树""" if start_index > end_index: return None # 将数组中间节点作为根节点 mid = (start_index + end_index) // 2 root = TreeNode(nums[mid]) # 递归建立左子树和右子树 root.left = self.build_tree( nums=nums, start_index=start_index, end_index=mid - 1 ) root.right = self.build_tree( nums=nums, start_index=mid + 1, end_index=end_index ) return root @property def root(self) -> TreeNode | None: return self.__root def search(self, num: int) -> TreeNode | None: """查找节点""" cur: TreeNode | None = self.__root # 循环查找,越过叶节点后跳出 while cur is not None: # 目标节点在 cur 的右子树中 if cur.val < num: cur = cur.right # 目标节点在 cur 的左子树中 elif cur.val > num: cur = cur.left # 找到目标节点,跳出循环 else: break return cur def insert(self, num: int) -> None: """插入节点""" # 若树为空,直接提前返回 if self.__root is None: return # 循环查找,越过叶节点后跳出 cur, pre = self.__root, None while cur is not None: # 找到重复节点,直接返回 if cur.val == num: return pre = cur # 插入位置在 cur 的右子树中 if cur.val < num: cur = cur.right # 插入位置在 cur 的左子树中 else: cur = cur.left # 插入节点 node = TreeNode(num) if pre.val < num: pre.right = node else: pre.left = node def remove(self, num: int) -> None: """删除节点""" # 若树为空,直接提前返回 if self.__root is None: return # 循环查找,越过叶节点后跳出 cur, pre = self.__root, None while cur is not None: # 找到待删除节点,跳出循环 if cur.val == num: break pre = cur # 待删除节点在 cur 的右子树中 if cur.val < num: cur = cur.right # 待删除节点在 cur 的左子树中 else: cur = cur.left # 若无待删除节点,则直接返回 if cur is None: return # 子节点数量 = 0 or 1 if cur.left is None or cur.right is None: # 当子节点数量 = 0 / 1 时, child = null / 该子节点 child = cur.left or cur.right # 删除节点 cur if pre.left == cur: pre.left = child else: pre.right = child # 子节点数量 = 2 else: # 获取中序遍历中 cur 的下一个节点 tmp: TreeNode = cur.right while tmp.left is not None: tmp = tmp.left # 递归删除节点 tmp self.remove(tmp.val) # 用 tmp 覆盖 cur cur.val = tmp.val """Driver Code""" if __name__ == "__main__": # 初始化二叉搜索树 nums = list(range(1, 16)) # [1, 2, ..., 15] bst = BinarySearchTree(nums=nums) print("\n初始化的二叉树为\n") print_tree(bst.root) # 查找节点 node = bst.search(7) print("\n查找到的节点对象为: {},节点值 = {}".format(node, node.val)) # 插入节点 bst.insert(16) print("\n插入节点 16 后,二叉树为\n") print_tree(bst.root) # 删除节点 bst.remove(1) print("\n删除节点 1 后,二叉树为\n") print_tree(bst.root) bst.remove(2) print("\n删除节点 2 后,二叉树为\n") print_tree(bst.root) bst.remove(4) print("\n删除节点 4 后,二叉树为\n") print_tree(bst.root)