# Write your MySQL query statement below
608. 树节点
给定一个表 tree，id 是树节点的编号， p_id 是它父节点的 id 。
+----+------+ | id | p_id | +----+------+ | 1 | null | | 2 | 1 | | 3 | 1 | | 4 | 2 | | 5 | 2 | +----+------+
树中每个节点属于以下三种类型之一：
写一个查询语句，输出所有节点的编号和节点的类型，并将结果按照节点编号排序。上面样例的结果为：
+----+------+ | id | Type | +----+------+ | 1 | Root | | 2 | Inner| | 3 | Leaf | | 4 | Leaf | | 5 | Leaf | +----+------+
解释
1
/ \
2 3
/ \
4 5
注意
如果树中只有一个节点，你只需要输出它的根属性。
原站题解
pythondata 解法, 执行用时: 522 ms, 内存消耗: 65.8 MB, 提交时间: 2024-05-27 11:30:25
import pandas as pd
def tree_node(tree: pd.DataFrame) -> pd.DataFrame:
# 复制输入 DataFrame 以避免修改原始数据
tree_copy = tree.copy()
# 使用自定义函数计算节点类型
def get_node_type(row: pd.DataFrame):
if pd.isnull(row['p_id']):
return 'Root'
elif row['id'] in tree_copy['p_id'].unique():
return 'Inner'
else:
return 'Leaf'
tree_copy['Type'] = tree_copy.apply(get_node_type, axis=1)
# 选择需要的列并按 id 排序
result_df = tree_copy[['id', 'Type']].sort_values(by='id')
return result_df
def tree_node2(tree: pd.DataFrame) -> pd.DataFrame:
a = tree
def tree_(x: pd.DataFrame):
if pd.isnull(x['p_id']):
return "Root"
elif tree['p_id'].isin([x['id']]).any(): # any() 等同或,有一个Turn就返回Ture
return 'Inner'
else:
return "Leaf"
a["type"] = tree.apply(tree_, axis=1)
return a[['id','type']]
mysql 解法, 执行用时: 490 ms, 内存消耗: 0 B, 提交时间: 2022-06-01 10:12:25
# Write your MySQL query statement below select id, case when p_id is null then 'Root' when id in (select distinct p_id from tree) then 'Inner' else 'Leaf' end as type from tree;
mysql 解法, 执行用时: 810 ms, 内存消耗: 0 B, 提交时间: 2022-05-27 15:01:30
# Write your MySQL query statement below select id, case when p_id is null then 'Root' when id in (select distinct p_id from tree) then 'Inner' else 'Leaf' end as type from tree;