[최원준] Day08

leetcode3/최원준/1886. Determine Whether Matrix Can Be Obtained By Rotation.py

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+#
+
+'''
+1. 아이디어 :
+-
+
+2. 시간복잡도 :
+    O(n * m)
+
+3. 자료구조/알고리즘 :
+-
+
+'''
+class Solution:
+    def findRotation(self, mat: List[List[int]], target: List[List[int]]) -> bool:
+        n = len(mat)
+        m = len(mat[0])
+        # 0, 90, 180, 270
+        ans = [1, 1, 1, 1]
+        for row in range(n):
+            for col in range(m):
+                if mat[row][col] != target[row][col]:
+                    ans[0] = 0
+                if mat[row][col] != target[col][n-row-1]:
+                    ans[1] = 0
+                if mat[row][col] != target[n-row-1][m-col-1]:
+                    ans[2] = 0
+                if mat[row][col] != target[n-col-1][row]:
+                    ans[3] = 0
+
+        return sum(ans) >= 1
+

leetcode3/최원준/894. All Possible Full Binary Trees.py

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+#
+
+'''
+1. 아이디어 :
+backtracking으로 왼쪽과 오른쪽 브랜치에 각각 몇개의 노드가 들어갈지 매번 계산한다.
+
+2. 시간복잡도 :
+    O(2^n)
+
+3. 자료구조/알고리즘 :
+backtracking
+
+'''
+# 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 allPossibleFBT(self, n: int) -> List[Optional[TreeNode]]:
+
+        # def copy_tree(original_node):
+        #     if not original_node:
+        #         return
+
+        #     copied = TreeNode(0)
+        #     copied.left = copy_tree(original_node.left)
+        #     copied.right = copy_tree(original_node.right)
+
+        #     return copied
+
+
+        def backtrack(nodes):
+
+            if nodes == 1: 
+                return [TreeNode(0)]
+
+            trees = []
+            for left_count in range(1, nodes-1, 2):
+                right_count = nodes - 1 - left_count
+
+                left_trees = backtrack(left_count)
+                right_trees = backtrack(right_count)
+
+                for left_tree in left_trees:
+                    for right_tree in right_trees:
+                        root = TreeNode(0)
+                        root.left = left_tree
+                        root.right = right_tree
+
+                        trees.append(root)
+            return trees
+
+        return backtrack(n)
+
+