124 lines
3.8 KiB
Python
124 lines
3.8 KiB
Python
"""
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二叉搜素树满足一下条件：
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1. 对于根节点，左子树中所有节点的值 小于 根节点的值 小于 右子树中所有节点的值。
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2. 任意节点的左、右子树也是二叉搜索树，即同样满足条件 1. 。
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查找节点：
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1. 从根节点查找，循环比较节点值和要查找的值大小
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2. 若节点值小于查找值，则走右子树
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3. 若节点值大于查找值，则走左子树
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"""
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from typing import Self
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class BinarySearchTree:
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def __init__(self, val: int = None, key: int = None):
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self.val: int | None = val
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self.key: int | None = key
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self.left: BinarySearchTree | None = None
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self.right: BinarySearchTree | None = None
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def search_val(self, key: int) -> int | None:
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if self.key is None:
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return None
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cur = self
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while cur:
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if cur.key == key:
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return cur.val
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if cur.key > key:
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cur = cur.left
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else:
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cur = cur.right
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return None
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def insert(self, key: int, val: int):
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if self.val is None:
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self.key = key
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self.val = val
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return
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cur, pre = self, None
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while cur:
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if cur.key == key:
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cur.val = val
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return
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if cur.key > key:
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if cur.left is None:
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cur.left = BinarySearchTree(key, val)
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return
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else:
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cur, pre = cur.left, cur
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else:
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if cur.right is None:
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cur.right = BinarySearchTree(key, val)
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return
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else:
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cur, pre = cur.right, cur
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if pre.key < key:
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pre.right = BinarySearchTree(key, val)
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else:
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pre.left = BinarySearchTree(key, val)
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def remove(self, key: int) -> int | None:
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if self.key is None:
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return None
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cur, pre = self, None
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while cur:
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if cur.key == key:
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break
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if cur.key > key:
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cur, pre = cur.left, cur
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else:
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cur, pre = cur.right, cur
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if cur is None:
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# 没有找到要删除的节点
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return
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if cur.left is None or cur.right is None:
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# 子节点数量有 0 个或 1 个
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child: BinarySearchTree | None = cur.left or cur.right
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if cur.key != self.key:
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# 不是根节点
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if pre.left.key == cur.key:
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# 查找到的当前节点是前一个节点的左子节点
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pre.left = child
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else:
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pre.right = child
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else:
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if child:
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self.key = child.key
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self.val = child.val
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else:
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self.key = None
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self.val = None
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else:
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# 子节点数量有两个, 无法直接删除它，而需要使用一个节点替换该节点
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# 由于要保持二叉搜索树“左子树 < 根节点 < 右子树”的性质，
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# 因此这个节点可以是右子树的最小节点或左子树的最大节点。
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temp, pre1 = cur.right, cur
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# 找右子树的最小节点
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while temp.left:
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temp, pre1 = temp.left, temp
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if pre.left.key == cur.key:
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pre.left = temp
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pre1.left = None
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else:
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pre.right = temp
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pre1.left = None
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def inorder(self, bst: Self):
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# 中序遍历的结果是升序
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if bst.left:
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self.inorder(self.left)
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print(self.val)
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if bst.right:
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self.inorder(self.right)
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