|
|
|
"""
|
|
|
|
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: int = (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) -> TreeNode | None:
|
|
|
|
"""插入节点"""
|
|
|
|
# 若树为空,直接提前返回
|
|
|
|
if self.__root is None:
|
|
|
|
return None
|
|
|
|
|
|
|
|
# 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 None
|
|
|
|
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 val
|
|
|
|
node = TreeNode(num)
|
|
|
|
if pre.val < num:
|
|
|
|
pre.right = node
|
|
|
|
else:
|
|
|
|
pre.left = node
|
|
|
|
return node
|
|
|
|
|
|
|
|
def remove(self, num: int) -> TreeNode | None:
|
|
|
|
"""删除节点"""
|
|
|
|
# 若树为空,直接提前返回
|
|
|
|
if self.__root is None:
|
|
|
|
return None
|
|
|
|
|
|
|
|
# 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
|
|
|
|
if cur.val < num: # 待删除节点在 cur 的右子树中
|
|
|
|
cur = cur.right
|
|
|
|
else: # 待删除节点在 cur 的左子树中
|
|
|
|
cur = cur.left
|
|
|
|
# 若无待删除节点,则直接返回
|
|
|
|
if cur is None:
|
|
|
|
return None
|
|
|
|
|
|
|
|
# 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
|
|
|
|
nex: TreeNode = self.get_inorder_next(cur.right)
|
|
|
|
tmp: int = nex.val
|
|
|
|
# 递归删除节点 nex
|
|
|
|
self.remove(nex.val)
|
|
|
|
# 将 nex 的值复制给 cur
|
|
|
|
cur.val = tmp
|
|
|
|
return cur
|
|
|
|
|
|
|
|
def get_inorder_next(self, root: TreeNode | None) -> TreeNode | None:
|
|
|
|
"""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)"""
|
|
|
|
if root is None:
|
|
|
|
return root
|
|
|
|
# 循环访问左子节点,直到叶节点时为最小节点,跳出
|
|
|
|
while root.left is not None:
|
|
|
|
root = root.left
|
|
|
|
return root
|
|
|
|
|
|
|
|
|
|
|
|
"""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
|
|
|
|
node = 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)
|