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