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‎Problemset/0968-Binary Tree Cameras/README.md

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# 968. Binary Tree Cameras
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You are given the `root` of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.
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Return *the minimum number of cameras needed to monitor all nodes of the tree*.
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#### Example 1:
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![](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png)
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<pre>
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<strong>Input:</strong> root = [0,0,null,0,0]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong> One camera is enough to monitor all nodes if placed as shown.
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</pre>
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#### Example 2:
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![](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png)
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<pre>
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<strong>Input:</strong> root = [0,0,null,0,null,0,null,null,0]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong> At least two cameras are needed to monitor all nodes of the tree. The above image shows one of the valid configurations of camera placement.
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</pre>
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#### Constraints:
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* The number of nodes in the tree is in the range `[1, 1000]`.
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* `Node.val == 0`
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## Solutions (Python)
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### 1. Solution
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```Python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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from functools import cache
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class Solution:
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@cache
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def minCameraCover(self, root: Optional[TreeNode]) -> int:
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if root is None:
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return 0
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ret = self.coverBySelf(root)
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if root.left is not None:
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ret = min(ret, self.coverBySelf(root.left) +
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self.minCameraCover(root.right))
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if root.right is not None:
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ret = min(ret, self.coverBySelf(root.right) +
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self.minCameraCover(root.left))
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return ret
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@cache
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def coverBySelf(self, root: TreeNode) -> int:
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ret = 1
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if root.left is not None:
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ret += self.coverByParent(root.left)
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if root.right is not None:
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ret += self.coverByParent(root.right)
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return ret
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@cache
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def coverByParent(self, root: TreeNode) -> int:
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return min(self.coverBySelf(root), self.minCameraCover(root.left) + self.minCameraCover(root.right))
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```

‎Problemset/0968-Binary Tree Cameras/README_CN.md

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# 968. 监控二叉树
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给定一个二叉树，我们在树的节点上安装摄像头。
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节点上的每个摄影头都可以监视**其父对象、自身及其直接子对象**。
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计算监控树的所有节点所需的最小摄像头数量。
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#### 示例 1:
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![](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_01.png)
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<pre>
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<strong>输入:</strong> root = [0,0,null,0,0]
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<strong>输出:</strong> 1
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<strong>解释:</strong> 如图所示，一台摄像头足以监控所有节点。
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</pre>
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#### 示例 2:
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![](https://assets.leetcode.com/uploads/2018/12/29/bst_cameras_02.png)
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<pre>
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<strong>输入:</strong> root = [0,0,null,0,null,0,null,null,0]
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<strong>输出:</strong> 2
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<strong>解释:</strong> 需要至少两个摄像头来监视树的所有节点。 上图显示了摄像头放置的有效位置之一。
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</pre>
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#### 提示:
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1. 给定树的节点数的范围是 `[1, 1000]`。
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2. 每个节点的值都是 0。
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## 题解 (Python)
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### 1. 题解
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```Python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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from functools import cache
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class Solution:
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@cache
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def minCameraCover(self, root: Optional[TreeNode]) -> int:
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if root is None:
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return 0
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ret = self.coverBySelf(root)
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if root.left is not None:
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ret = min(ret, self.coverBySelf(root.left) +
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self.minCameraCover(root.right))
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if root.right is not None:
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ret = min(ret, self.coverBySelf(root.right) +
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self.minCameraCover(root.left))
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return ret
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@cache
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def coverBySelf(self, root: TreeNode) -> int:
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ret = 1
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if root.left is not None:
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ret += self.coverByParent(root.left)
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if root.right is not None:
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ret += self.coverByParent(root.right)
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return ret
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@cache
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def coverByParent(self, root: TreeNode) -> int:
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return min(self.coverBySelf(root), self.minCameraCover(root.left) + self.minCameraCover(root.right))
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```

‎Problemset/0968-Binary Tree Cameras/Solution.py

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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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from functools import cache
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class Solution:
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@cache
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def minCameraCover(self, root: Optional[TreeNode]) -> int:
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if root is None:
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return 0
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ret = self.coverBySelf(root)
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if root.left is not None:
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ret = min(ret, self.coverBySelf(root.left) +
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self.minCameraCover(root.right))
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if root.right is not None:
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ret = min(ret, self.coverBySelf(root.right) +
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self.minCameraCover(root.left))
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return ret
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@cache
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def coverBySelf(self, root: TreeNode) -> int:
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ret = 1
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if root.left is not None:
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ret += self.coverByParent(root.left)
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if root.right is not None:
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ret += self.coverByParent(root.right)
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return ret
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@cache
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def coverByParent(self, root: TreeNode) -> int:
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return min(self.coverBySelf(root), self.minCameraCover(root.left) + self.minCameraCover(root.right))

‎README.md

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[965][965l] |[Univalued Binary Tree][965] |![py]
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[966][966l] |[Vowel Spellchecker][966] |![py]
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[967][967l] |[Numbers With Same Consecutive Differences][967] |![rb]&nbsp;&nbsp;![rs]
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[968][968l] |[Binary Tree Cameras][968] |![py]
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[969][969l] |[Pancake Sorting][969] |![py]
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[970][970l] |[Powerful Integers][970] |![rs]
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[973][973l] |[K Closest Points to Origin][973] |![rs]
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[965]:Problemset/0965-Univalued%20Binary%20Tree/README.md#965-univalued-binary-tree
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[966]:Problemset/0966-Vowel%20Spellchecker/README.md#966-vowel-spellchecker
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[967]:Problemset/0967-Numbers%20With%20Same%20Consecutive%20Differences/README.md#967-numbers-with-same-consecutive-differences
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[968]:Problemset/0968-Binary%20Tree%20Cameras/README.md#968-binary-tree-cameras
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[969]:Problemset/0969-Pancake%20Sorting/README.md#969-pancake-sorting
21712173
[970]:Problemset/0970-Powerful%20Integers/README.md#970-powerful-integers
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[973]:Problemset/0973-K%20Closest%20Points%20to%20Origin/README.md#973-k-closest-points-to-origin
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[965l]:https://leetcode.com/problems/univalued-binary-tree/
37053707
[966l]:https://leetcode.com/problems/vowel-spellchecker/
37063708
[967l]:https://leetcode.com/problems/numbers-with-same-consecutive-differences/
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[968l]:https://leetcode.com/problems/binary-tree-cameras/
37073710
[969l]:https://leetcode.com/problems/pancake-sorting/
37083711
[970l]:https://leetcode.com/problems/powerful-integers/
37093712
[973l]:https://leetcode.com/problems/k-closest-points-to-origin/

‎README_CN.md

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[965][965l] |[单值二叉树][965] |![py]
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[966][966l] |[元音拼写检查器][966] |![py]
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[967][967l] |[连续差相同的数字][967] |![rb]&nbsp;&nbsp;![rs]
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[968][968l] |[监控二叉树][968] |![py]
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[969][969l] |[煎饼排序][969] |![py]
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[970][970l] |[强整数][970] |![rs]
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[973][973l] |[最接近原点的 K 个点][973] |![rs]
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[965]:Problemset/0965-Univalued%20Binary%20Tree/README_CN.md#965-单值二叉树
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[966]:Problemset/0966-Vowel%20Spellchecker/README_CN.md#966-元音拼写检查器
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[967]:Problemset/0967-Numbers%20With%20Same%20Consecutive%20Differences/README_CN.md#967-连续差相同的数字
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[968]:Problemset/0968-Binary%20Tree%20Cameras/README_CN.md#968-监控二叉树
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[969]:Problemset/0969-Pancake%20Sorting/README_CN.md#969-煎饼排序
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[970]:Problemset/0970-Powerful%20Integers/README_CN.md#970-强整数
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[973]:Problemset/0973-K%20Closest%20Points%20to%20Origin/README_CN.md#973-最接近原点的-K-个点
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[965l]:https://leetcode.cn/problems/univalued-binary-tree/
37053707
[966l]:https://leetcode.cn/problems/vowel-spellchecker/
37063708
[967l]:https://leetcode.cn/problems/numbers-with-same-consecutive-differences/
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[968l]:https://leetcode.cn/problems/binary-tree-cameras/
37073710
[969l]:https://leetcode.cn/problems/pancake-sorting/
37083711
[970l]:https://leetcode.cn/problems/powerful-integers/
37093712
[973l]:https://leetcode.cn/problems/k-closest-points-to-origin/

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