Merge pull request #1605 from shinsj4653/main

[shinsj4653] Week 12 Solutions

Author: shinsj4653

non-overlapping-intervals/shinsj4653.py

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+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+

number-of-connected-components-in-an-undirected-graph/shinsj4653.py

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+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+
+

remove-nth-node-from-end-of-list/shinsj4653.py

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+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+

same-tree/shinsj4653.py

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+"""
+[문제풀이]
+# Inputs
+두 이진 트리의 노드 배열들 p, q
+
+# Outputs
+두 트리가 같은지 다른지 체크
+
+# Constraints
+- The number of nodes in both trees is in the range [0, 100].
+- -104 <= Node.val <= 104
+
+# Ideas
+둘 다 bfs?? 때리면 될 것 같은데?
+동시에 탐색하면서 다른 노드 나오면 바로 종료
+
+[회고]
+풀긴 풀었는데 좀 더 깔금한 풀이가 있을까?
+->
+
+"""
+
+# 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 isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        def dfs(p_tree, q_tree):
+            print('p: ', p_tree)
+            print('q: ', q_tree)
+
+            if p_tree is not None and q_tree is not None and p_tree.val == q_tree.val:
+                print('add left and right')
+                return dfs(p_tree.left, q_tree.left) and dfs(p_tree.right, q_tree.right)
+
+            if (p_tree == q_tree == None):
+                return True
+
+            if (p_tree is not None and q_tree is None) or \
+                    (p_tree is None and q_tree is not None) or \
+                    (p_tree is not None and q_tree is not None and p_tree.val != q_tree.val):
+                print("not same!!")
+                return False
+
+        return dfs(p, q)
+
+# 해설
+class Solution:
+    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        if p is None and q is None:
+            return True
+        if p is None or q is None or p.val != q.val:
+            return False
+        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
+
+
+
+
+

serialize-and-deserialize-binary-tree/shinsj4653.py

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+"""
+[문제풀이]
+# Inputs
+
+# Outputs
+
+# Constraints
+
+# Ideas
+
+[회고]
+
+"""
+
+
+