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437. Path Sum III # 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 pathSum(self, root: Optional[TreeNode], targetSum: int) -> int: count =0 def dfs(node, just_started, sub_sum): nonlocal count if node =..

1372. Longest ZigZag Path in a Binary Tree # 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 longestZigZag(self, root: Optional[TreeNode]) -> int: res = 0 def dfs(node, left2right, depth): nonlocal res ..

1448. Count Good Nodes in Binary Tree https://youtu.be/7cp5imvDzl4?si=6kO3CQDgpb403ogg # 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 goodNodes(self, root: TreeNode) -> int: def dfs(node, maxVal): """ ..

872. Leaf-Similar Trees 내 풀이# 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 getleaf(self, root): leaf1 = [] s1 = [(root, 1)] # node, depth max_depth = 1 while s1: node, depth = s1.pop() ..

2130. Maximum Twin Sum of a Linked List Reverse 뒤집은 노드 만들기curr = head # 원래 리스트의 시작prev = None # 뒤집힌 리스트의 시작 (처음엔 없음)while curr: next = curr.next # 다음 노드 기억 curr.next = prev # 현재 노드의 방향을 뒤집음 prev = curr # prev를 현재 노드로 업데이트 curr = next # 다음 노드로 이동 이 루프가 끝나면:curr은 None (즉, 리스트 끝을 넘은 상태)prev는 뒤집힌 리스트의 시작 지점, 즉 새로운 head!🔍 예를 들어서:리스트가 1 → 2 → 3 ..

206. Reverse Linked List https://leetcode.com/problems/reverse-linked-list/description/  나도 stack 으로 풀었으나, 제대로 못푼 느낌??# Definition for singly-linked list.# class ListNode:# def __init__(self, val=0, next=None):# self.val = val# self.next = nextclass Solution: def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: # 스텍으로 푸는 법 stack = [] ..

328. Odd Even Linked List # Definition for singly-linked list.# class ListNode:# def __init__(self, val=0, next=None):# self.val = val# self.next = nextclass Solution: def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]: # 노드가 2개 까지는 그대로 반환 if not head or not head.next or not head.next.next: return head odd = current = hea..

2095. Delete the Middle Node of a Linked List # Definition for singly-linked list.# class ListNode:# def __init__(self, val=0, next=None):# self.val = val# self.next = nextclass Solution: def deleteMiddle(self, head: Optional[ListNode]) -> Optional[ListNode]: if not head or not head.next: return None one_step = head two_step = head.next.next..

649. Dota2 Senate class Solution: def predictPartyVictory(self, senate: str) -> str: r_q = deque() d_q = deque() for i, s in enumerate(senate): if s == "R": r_q.append(i) else: d_q.append(i) while r_q and d_q: d = d_q.popleft() r = r_q.popleft() if d  https://youtu.be/zZA5KskfMuQ?s..

933. Number of Recent Calls import collectionsclass RecentCounter: def __init__(self): self.q = collections.deque() self.q_len = 0 def ping(self, t: int) -> int: self.q.append(t) self.q_len +=1 while self.q[0]