Merge branch 'main' of https://github.com/hu6r1s/leetcode-study

Author: hu6r1s

maximum-depth-of-binary-tree/hu6r1s.py

@@ -0,0 +1,50 @@
+from collections import deque
+
+# 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 maxDepth(self, root: Optional[TreeNode]) -> int:
+        if not root:
+            return 0
+
+        def bfs(n):
+            queue = deque([n])
+            depth = 0
+
+            while queue:
+                depth += 1
+                for _ in range(len(queue)):
+                    node = queue.popleft()
+                    if node.left:
+                        queue.append(node.left)
+                    if node.right:
+                        queue.append(node.right)
+            return depth
+
+        def dfs(n):
+            stack = [n]
+            max_depth = 0
+            visited = {n: 1}
+
+            while stack:
+                node = stack.pop()
+                depth = visited[node]
+                max_depth = max(max_depth, depth)
+                if node.left:
+                    visited[node.left] = depth + 1
+                    stack.append(node.left)
+                if node.right:
+                    visited[node.right] = depth + 1
+                    stack.append(node.right)
+            return max_depth
+        
+        return dfs(root)
+
+
+"""
+bfs 방식으로 left나 right가 있으면 스택에 넣고 depth + 1, dfs보다 비효율인 듯
+"""

merge-two-sorted-lists/hu6r1s.py

@@ -0,0 +1,26 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    # 시간복잡도: O(n + m)  -> 두 리스트의 모든 노드를 한 번씩 방문
+    # 공간복잡도: O(1)      -> 기존 노드 재배치, 추가 메모리 거의 없음
+    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+        merged_list = ListNode()
+        tail = merged_list
+
+        while list1 and list2:
+            if list1.val < list2.val:
+                tail.next = list1
+                list1 = list1.next
+            else:
+                tail.next = list2
+                list2 = list2.next
+            tail = tail.next
+        
+        if list1:
+            tail.next = list1
+        else:
+            tail.next = list2
+        return merged_list.next