小黑开始了拉歌训练，第一次进入部室馆，被通知要去当主持人心里有些紧张的leetcode之旅:337. 打家劫舍 III

news2025/1/12 10:03:02

小黑代码（小黑卡在了bug中，上午一步步探索做出，非常NB!!!）

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rob(self, root: Optional[TreeNode]) -> int:
        # 递归函数进行动态规划
        def dfs(node):
            # 节点为空
            if not node:
                return 0, 0, 0
            # 叶子结点
            if not (node.left or node.right):
                return 0, 0, node.val
            # 动态规划操作
            first_1, second_1, node1_val = dfs(node.left)
            first_2, second_2, node2_val = dfs(node.right)
            third_1 = max(second_1, first_1+node1_val)
            third_2 = max(second_2, first_2+node2_val)
            return second_1+second_2, third_1+third_2, node.val
        first_root, second_root, val = dfs(root)
        return max(second_root, first_root+val)

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动态规划法

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rob(self, root: Optional[TreeNode]) -> int:
        # 递归函数
        # 返回[不选择该结点获得收益, 选择该结点获得的收益]
        def dfs(node):
            # 结点为空
            if not node:
                return [0, 0]
            # 获得左右结点的数组
            l = dfs(node.left)
            r = dfs(node.right)
            # 不选择该结点
            not_select = max(l[0], l[1]) + max(r[0], r[1])
            # 选择该结点
            select = node.val + l[0] + r[0]
            return [not_select, select]
        return max(dfs(root))

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数据库练习

608. 树节点

使用union

# Write your MySQL query statement below
SELECT
    id, 'Root' AS 'type'
FROM
    Tree
WHERE
    p_id IS NULL

UNION

SELECT
    id, 'Leaf' AS 'type'
FROM
    Tree
WHERE
    id NOT IN
        (SELECT
            p_id
        FROM
            Tree
        WHERE
            p_id IS NOT NULL
        )
    AND p_id IS NOT NULL
UNION

SELECT
    id, 'Inner' AS 'type'
FROM
    Tree
WHERE
    id IN 
        (SELECT DISTINCT
            p_id
        FROM
            Tree
        WHERE
            p_id IS NOT NULL
        )
    AND p_id IS NOT NULL

Case法（小黑第一次使用，了解了sql中的case的基本用法）

# Write your MySQL query statement below
SELECT
    id,
    CASE
        WHEN tree.p_id IS NULL
            THEN 'Root'
        WHEN tree.id IN (SELECT DISTINCT p_id FROM Tree WHERE p_id IS NOT NULL) AND tree.p_id IS NOT NULL
            THEN 'Inner'
        WHEN tree.id NOT IN (SELECT DISTINCT p_id FROM TREE WHERE p_id IS NOT NULL)
            THEN 'Leaf'
        ELSE 'other'
    END AS Type
FROM
    Tree

IF分支法

# Write your MySQL query statement below
SELECT
    id,
    IF(ISNULL(Tree.p_id), 'Root', IF(Tree.id IN (SELECT DISTINCT p_id FROM Tree WHERE p_id IS NOT NULL) AND Tree.p_id IS NOT NULL, 'Inner', 'Leaf')) AS 'type'
FROM
    Tree