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			160 lines
		
	
	
		
			4.5 KiB
		
	
	
	
		
			Python
		
	
	
	
	
	
			
		
		
	
	
			160 lines
		
	
	
		
			4.5 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) -> 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
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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
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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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    def remove(self, num: int) -> 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
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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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            # 待删除节点在 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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            # 待删除节点在 cur 的左子树中
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            else:
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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
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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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            tmp: TreeNode = cur.right
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            while tmp.left is not None:
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                tmp = tmp.left
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            # 递归删除节点 tmp
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            self.remove(tmp.val)
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            # 用 tmp 覆盖 cur
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            cur.val = tmp.val
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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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    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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