feat: Solve binary-tree-level-order-traversal problem

Author: hu6r1s

binary-tree-level-order-traversal/hu6r1s.py

@@ -0,0 +1,27 @@
+# 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
+from collections import deque
+
+class Solution:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        if not root:
+            return []
+
+        queue = deque([root])
+        result = []
+        while queue:
+            tmp = []
+            for _ in range(len(queue)):
+                node = queue.popleft()
+                tmp.append(node.val)
+
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+            result.append(tmp)
+        return result

graph-valid-tree/hu6r1s.py

@@ -0,0 +1,18 @@
+if len(edges) != n - 1:
+            return False
+
+        graph = [[] for _ in range(n)]
+        for node, adj in edges:
+            graph[node].append(adj)
+            graph[adj].append(node)
+
+        visited = set()
+
+        def dfs(node):
+            visited.add(node)
+            for adj in graph[node]:
+                if adj not in visited:
+                    dfs(adj)
+
+        dfs(0)
+        return len(visited) == n

number-of-islands/hu6r1s.py

@@ -0,0 +1,36 @@
+from collections import deque
+
+class Solution:
+    """
+    문제를 보니 바로 그래프 탐색이 떠올라서 bfs 알고리즘을 사용해서 구현
+    백준 문제에서 많이 풀어보던건데 너무 오래 되어 계속 헷갈렸음
+    다시 공부해야함
+    """
+    def numIslands(self, grid: List[List[str]]) -> int:        
+        def bfs(grid, i, j):
+            queue = deque()
+            queue.append([i, j])
+            grid[i][j] = "0"
+            while queue:
+                x, y = queue.popleft()
+                for k in range(4):
+                    nx = x + dx[k]
+                    ny = y + dy[k]
+                    if nx < 0 or nx >= n or ny < 0 or ny >= m:
+                        continue
+                    if grid[nx][ny] == "0":
+                        continue
+                    grid[nx][ny] = "0"
+                    queue.append([nx, ny])
+            
+
+        dx = [-1, 1, 0, 0]
+        dy = [0, 0, -1, 1]
+        n, m = len(grid), len(grid[0])
+        cnt = 0
+        for i in range(n):
+            for j in range(m):
+                if grid[i][j] == "1":
+                    bfs(grid, i, j)
+                    cnt += 1
+        return cnt