[shinsj4653] Week 15 Solutions

alien-dictionary/shinsj4653.py

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

construct-binary-tree-from-preorder-and-inorder-traversal/shinsj4653.py

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

longest-palindromic-substring/shinsj4653.py

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

rotate-image/shinsj4653.py

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

subtree-of-another-tree/shinsj4653.py

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+"""
+[문제풀이]
+# Inputs
+root, subRoot
+
+# Outputs
+subRoot의 트리가 root 트리의 서브 트리인지에 대한 여부
+
+# Constraints
+- The number of nodes in the root tree is in the range [1, 2000].
+- The number of nodes in the subRoot tree is in the range [1, 1000].
+- -10^4 <= root.val <= 10^4
+- -10^4 <= subRoot.val <= 10^4
+
+# Ideas
+정확히 구조랑 노드 개수도 같아야 함
+-> subRoot의 root 노드 찾으면, 탐색 시작
+-> 탐색 끝나고 나서의 node 수 반환
+탐색 도중에 node.value 다르면 false 반환
+
+[회고]
+
+"""
+# 1차 제출 코드
+# 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 isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+
+        def compare(r, sr):
+            print('start compare')
+            if r is not None and sr.val is not None:
+                if r.val != sr.val:
+                    print('not same, exit compare')
+                    return False
+
+                elif r.val == sr.val == None:
+                    print('same, exit compare')
+                    return True
+
+                else:
+                    print('continue compare')
+                    return compare(r.left, sr.left) or compare(r.right, sr.right)
+
+            return True
+
+        def dfs(cur):
+            if cur.val != subRoot.val:
+                print('going left in root')
+                dfs(cur.left)
+
+                print('going right in root')
+                dfs(cur.right)
+
+            else:
+                print('begin compare')
+                return compare(cur, subRoot)
+            return
+
+        return dfs(root)
+
+# gpt 수정 코드
+
+class Solution:
+    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+
+        def isSameTree(s, t):
+            if not s and not t:
+                return True
+            if not s or not t:
+                return False
+            if s.val != t.val:
+                return False
+            return isSameTree(s.left, t.left) and isSameTree(s.right, t.right)
+
+        def dfs(node):
+            if not node:
+                return False
+            if isSameTree(node, subRoot):
+                return True
+            # 아래 둘 중 하나라도 True면 즉시 return True
+            return dfs(node.left) or dfs(node.right)
+
+        return dfs(root)
+
+