[LeetCode] Customes Who Never Order (is null, join)
기본적으로 Customers join Orders 구조 select * from customers c join orders o on c.ID = o.CustomerID 이 상태로 출력해보면 "headers": ["Id", "Name", "Id", "CustomerId"], "values": [[1, "Joe", 2, 1], [3, "Sam", 1, 3]] 이렇게 나온다. 따라서 left join이 필요하다. select * from customers c left join orders o on c.ID = o.CustomerID left join의 결과는 {"headers": ["Id", "Name", "Id", "CustomerId"], "values": [[1, "Joe", 2, 1], [2, "Hen..
2021. 4. 9.
[LeetCode] Binary Tree Paths 파이썬 (Binary Tree, LCA, DFS)
class Solution(object): # @param {TreeNode} root # @return {string[]} def binaryTreePaths(self, root): result, path = [], [] self.binaryTreePathsRecu(root, path, result) return result def binaryTreePathsRecu(self, node, path, result): if node is None: return if node.left is node.right is None: ans = "" for n in path: ans += str(n.val) + "->" result.append(ans + str(node.val)) if node.left: path...
2021. 4. 7.