From a0076ed7fb3d6c7d5410d4f960c6a997d308d054 Mon Sep 17 00:00:00 2001 From: TORUS <42745810+TORUS0818@users.noreply.github.com> Date: Thu, 25 Jul 2024 10:24:14 +0900 Subject: [PATCH] Create answer.md --- easy/108/answer.md | 260 +++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 260 insertions(+) create mode 100644 easy/108/answer.md diff --git a/easy/108/answer.md b/easy/108/answer.md new file mode 100644 index 0000000..b810aa6 --- /dev/null +++ b/easy/108/answer.md @@ -0,0 +1,260 @@ +# Step1 + +かかった時間：5min + +計算量：nums.length=Nとして、 + +時間計算量：O(N) + +空間計算量：O(NlogN) + +recursive-DFS +```python +# 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 sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: + if not nums: + return None + + root_value_index = len(nums) // 2 + root = TreeNode(val = nums[root_value_index]) + root.left = self.sortedArrayToBST(nums[:root_value_index]) + root.right = self.sortedArrayToBST(nums[root_value_index + 1:]) + + return root +``` +思考ログ： +- 直近で解き方を見ている +- 配列がソートされているので真ん中で割って根から順に木を生やせば良い +- 例によって再帰なのでスタックを気にしておく + - 今回はlog(10^4)なので問題なさそう +- 配列のスライスがうまく機能しているか（漏れたりしてないか）少し脳内テスト + +loop-DFS +```python +# 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 sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: + root = TreeNode() + node_nums_pairs = [(root, nums)] + while node_nums_pairs: + node, subset_nums = node_nums_pairs.pop() + mid_index = len(subset_nums) // 2 + node.val = subset_nums[mid_index] + + left_nums = subset_nums[:mid_index] + if left_nums: + node.left = TreeNode() + node_nums_pairs.append((node.left, left_nums)) + right_nums = subset_nums[mid_index + 1:] + if right_nums: + node.right = TreeNode() + node_nums_pairs.append((node.right, right_nums)) + + return root +``` +思考ログ： +- 配列を連れ回しているのは良くないか + - インデックスで管理すればいいよねという話 + +# Step2 + +講師役目線でのセルフツッコミポイント： +- 特に思いつかなかった + +参考にした過去ログなど： +- https://github.com/Yoshiki-Iwasa/Arai60/pull/28 + - midの取り方（left or right）に気が回っていなかった +- https://github.com/kazukiii/leetcode/pull/25 + - 確かに先に木を作る方法もあった + > dummyの値を持ったcomplete binary treeを構築して、inorder traversalしながら値をセットしても良さそう + - ボックス化 + - https://ja.wikipedia.org/wiki/%E3%83%9C%E3%83%83%E3%82%AF%E3%82%B9%E5%8C%96 +- https://github.com/fhiyo/leetcode/pull/26 +- https://github.com/Mike0121/LeetCode/pull/13 +- https://github.com/hayashi-ay/leetcode/pull/29 + - ２分木でもsentinel + - https://github.com/hayashi-ay/leetcode/pull/29/files#r1593455812 +- https://github.com/sakupan102/arai60-practice/pull/25 +- https://github.com/rossy0213/leetcode/pull/13 +- https://discord.com/channels/1084280443945353267/1183683738635346001/1209195844511858688 + +recursive-DFS（インデックス処理, 右半開区間ver） +```python +# 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 sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: + def build_bst(left: int, right: int): + if left >= right: + return None + + mid = (left + right) // 2 + node = TreeNode(nums[mid]) + node.left = build_bst(left, mid) + node.right = build_bst(mid + 1, right) + return node + + return build_bst(0, len(nums)) +``` +思考ログ： +- 一つ関数を定義する必要があるが、インデックスで処理できるからエコな実装になる + +recursive-DFS（インデックス処理, 閉区間ver） +```python +# 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 sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: + def build_bst(left: int, right: int): + if left > right: + return None + + mid = (left + right) // 2 + node = TreeNode(nums[mid]) + node.left = build_bst(left, mid - 1) + node.right = build_bst(mid + 1, right) + return node + + return build_bst(0, len(nums) - 1) +``` +思考ログ： +- 右半開区間の方が分かりやすいか + +親と子の情報を積んでいくやり方 +```python +# 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 sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: + left, mid, right = 0, len(nums) // 2, len(nums) + root = TreeNode(nums[mid]) + # [(parent_node, index_range_for_child, is_left_child)] + parent_with_child_infos = [ + (root, (left, mid), True), + (root, (mid + 1, right), False) + ] + while parent_with_child_infos: + parent_node, (left, right), is_left = parent_with_child_infos.pop() + if left >= right: + continue + mid = (left + right) // 2 + child_node = TreeNode(nums[mid]) + if is_left: + parent_node.left = child_node + else: + parent_node.right = child_node + parent_with_child_infos.append((child_node, (left, mid), True)) + parent_with_child_infos.append((child_node, (mid + 1, right), False)) + + return root +``` +思考ログ： +- 積む情報が多い、スタックの問題もあるが、再帰は簡潔に書けて良い + +先に木を作ってin-orderで値を埋めていく方法 +```python +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 sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: + # complete binary tree + def build_cbt(index: int): + if index >= len(nums): + return None + root = TreeNode() + root.left = build_cbt(2 * index + 1) + root.right = build_cbt(2 * index + 2) + return root + + nums_queue = deque(nums) + def set_values(cbt_root: Optional[TreeNode]) -> None: + nodes = [cbt_root] + while nodes: + node = nodes.pop() + if not node: + continue + set_values(node.left) + node.val = nums_queue.popleft() + set_values(node.right) + + cbt_root = build_cbt(0) + set_values(cbt_root) + + return cbt_root +``` +思考ログ： +- in-orderを書く機会が少ないのでせっかくなので書いておく + - https://github.com/kazukiii/leetcode/pull/25/files +- nonlocalを使わない形に少しアレンジ + +# Step3 + +かかった時間：3min + +```python +# 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 sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: + if not nums: + return None + + root = TreeNode() + node_range_pairs = [(root, (0, len(nums)))] + while node_range_pairs: + node, (left, right) = node_range_pairs.pop() + mid = (left + right) // 2 + node.val = nums[mid] + + if left < mid: + node.left = TreeNode() + node_range_pairs.append((node.left, (left, mid))) + if mid + 1 < right: + node.right = TreeNode() + node_range_pairs.append((node.right, (mid + 1, right))) + + return root +``` +思考ログ： +- 再帰でいい気がするが、スタック版を練習も兼ねて + +# Step4 + +```python +``` +思考ログ： +