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94. Binary Tree Inorder Traversal # 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 = rightclass Solution: def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]: # inorder: dfs 에서,.. 왼쪽, 현재, 오른쪽 순회 # dfs 니깐, stack 이나 재귀로 if not root:..

Inorder Traversal 순회왼쪽 -> 현재 -> 오른쪽https://youtu.be/hUaKh73y9lc Preorder Traversal 순회현재 -> 왼쪽 -> 오른쪽  https://youtu.be/KReQI3Muebg Postorder Traversal 순회왼쪽 -> 오른쪽 -> 현재   https://youtu.be/bJoMy7V0jQw

BST 값 추가 # 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 = rightclass Solution: def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]: def insert(node, val): if not node: # 노드 없으면, return TreeNod..

111. Minimum Depth of Binary Tree if not root: 루트 비어 있는 경우    return 0 : 깊이는 0 root 부터, 깊이는 1 자식이 둘다 없으면 leaf 노드 임if not node.left and not node.right:   return 깊이 최소 값 업데이트 # 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 = rightclass Solution: def minDepth(self, r..

2218. Maximum Value of K Coins From Piles https://youtu.be/ZRdEd_eun8g?si=nPvl34N3ezdyQO58 백트래킹 솔루션 : 테스트는 통과하나, 시간 제한에 걸림class Solution: def maxValueOfCoins(self, piles: List[List[int]], k: int) -> int: n = len(piles) def dfs(i, coins): if i == n: # 범위 벗어나서. i 는 piles 의 index return 0 res = dfs(i+1, coins) # skip current pile curP..

1220. Count Vowels Permutation https://youtu.be/VUVpTZVa7Ls?si=ebIzPzi9iQeVkNQa class Solution: def countVowelPermutation(self, n: int) -> int: """ Permutation: 순열 / Permutation 아래 규칙을 따르는 n 길이의 순열 개수 계산: Calculate the number of permutations of length `n` following these rules: - 문자열은 a, e, i, o, u로만 구성 / The string consists of only 'a', 'e', 'i', 'o', 'u'. ..

265. Paint House II class Solution: def minCostII(self, costs: List[List[int]]) -> int: N = len(costs) K = len(costs[0]) for n in range(N-1-1, -1, -1): for k in range(K): costs[n][k] += min([costs[n+1][i] for i in range(K) if i != k]) return min(costs[0])

256. Paint House 내 풀이 (맞춤)class Solution: def minCost(self, costs: List[List[int]]) -> int: n = len(costs) dp = [[0]*3 for _ in range(n)] dp[-1][:] = costs[-1][:] if n > 1: for r in range(n-1-1, -1, -1): for c in range(3): dp[r][c%3] = min(dp[r+1][(c+1)%3], dp[r+1][(c+2)%3]) + costs[r][c%3] #print(dp) ..

983. Minimum Cost For Tickets https://youtu.be/4pY1bsBpIY4?si=PbuLW3CZC5vqdJDg class Solution: def mincostTickets(self, days: List[int], costs: List[int]) -> int: dp = {} def dfs(i): if i == len(days): return 0 if i in dp: return dp[i] dp[i] = float("inf") # d: # of days, c: cost for d, c in zip..

91. Decode Ways https://youtu.be/6aEyTjOwlJU?si=gOHrfB_eRCdonSuR class Solution: def numDecodings(self, s: str) -> int: dp = {len(s) : 1} def dfs(i): if i in dp: return dp[i] if s[i] == "0": return 0 res = dfs(i+1) if (i+1  class Solution: def numDecodings(self, s: str) -> int: dp = {len(s) : 1} ..