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Original file line number Diff line number Diff line change
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### DFS(再帰)

時間計算量: O(N)<br>
空間計算量: O(N)<br>

---
### 1回目 (20m15s)
二分探索のアイデアでいけば書けることはパッと思いついたが、コードにうまくできず時間がかかった。
以前といた時よりはかなりアイデアそのものは簡単に感じた。
root.left, root.rightを作って、下から木を構築していく問題がやや苦手だが、今回の問題でかなり克服できたと考える。
(結果を伝播させていく問題はわかってきた。)


```py
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
if not nums:
return []

def sorted_array_to_bst_helper(left, right):
if left > right:
return

mid_index = (left + right) // 2
root = TreeNode(nums[mid_index])

root.left = sorted_array_to_bst_helper(left, mid_index - 1)
root.right = sorted_array_to_bst_helper(mid_index + 1, right)

return root

return sorted_array_to_bst_helper(0, len(nums) - 1)
```


### 2回目
left, mid, rightは、それぞれ_indexをつけるかつけないか統一した方が良いと考えた。
```python
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:

def sorted_array_to_bst_helper(left, right):
if left > right:
return
mid = (left + right) // 2
root = TreeNode(nums[mid])
root.left = sorted_array_to_bst_helper(left, mid - 1)
root.right = sorted_array_to_bst_helper(mid + 1, right)
return root

return sorted_array_to_bst_helper(0, len(nums) - 1)
```

### 3回目 (半開区間に変更)
```python
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
def sorted_array_to_bst_helper(left, right):
if left == right:
return
mid = (left + right) // 2
root = TreeNode(nums[mid])
root.left = sorted_array_to_bst_helper(left, mid)
root.right = sorted_array_to_bst_helper(mid + 1, right)
return root

return sorted_array_to_bst_helper(0, len(nums))
```


### helper関数なし
---
```python
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
if not nums:
return None
mid = len(nums) // 2
root = TreeNode(nums[mid])
root.left = self.sortedArrayToBST(nums[:mid])
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スライスはコピーが発生したかと思います。

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odaさん、コメントありがとうございます。
解釈間違っているかもしれませんが、
時間計算量: O(nlogn) ということでしょうか?

root.right = self.sortedArrayToBST(nums[mid + 1:])

return root
```

### DFS (スタック)

時間計算量: O(N)<br>
空間計算量: O(N)<br>

再帰をベースに考えて15分ほどで書いた。
---
```python
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
if not nums:
return []

mid = (len(nums) - 1) // 2
root = TreeNode(nums[mid])
node_and_divided_array = [(root, nums[:mid], nums[mid+1:])]
while node_and_divided_array:
node, nums_left, nums_right = node_and_divided_array.pop()

if nums_right:
mid_right = (len(nums_right) - 1) // 2
node.right = TreeNode(nums_right[mid_right])
node_and_divided_array.append((node.right, nums_right[:mid_right], nums_right[mid_right+1:]))

if nums_left:
mid_left = (len(nums_left) - 1) // 2
node.left = TreeNode(nums_left[mid_left])
node_and_divided_array.append((node.left, nums_left[:mid_left], nums_left[mid_left+1:]))

return root
```


### BFS

時間計算量: O(N)<br>
空間計算量: O(N)<br>

---
```python
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
if not nums:
return []

mid = (len(nums) - 1) // 2
root = TreeNode(nums[mid])
node_and_divided_array = deque([(root, nums[:mid], nums[mid+1:])])
while node_and_divided_array:
node, nums_left, nums_right = node_and_divided_array.popleft()

if nums_left:
mid_left = (len(nums_left) - 1) // 2
node.left = TreeNode(nums_left[mid_left])
node_and_divided_array.append((node.left, nums_left[:mid_left], nums_left[mid_left+1:]))

if nums_right:
mid_right = (len(nums_right) - 1) // 2
node.right = TreeNode(nums_right[mid_right])
node_and_divided_array.append((node.right, nums_right[:mid_right], nums_right[mid_right+1:]))

return root
```