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.
Return the minimum number of cameras needed to monitor all nodes of the tree.
Example 1:
o
/
C
/ \
o o
Input: root = [0,0,null,0,0]
Output: 1
Explanation: One camera is enough to monitor all nodes if placed as shown.
Example 2:
o
/
C
/
o
/
C
\
o
Input: root = [0,0,null,0,null,0,null,null,0]
Output: 2
Explanation: 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.
Constraints:
- The number of nodes in the tree is in the range
[1, 1000].
Node.val == 0
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# 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 minCameraCover(self, root: Optional[TreeNode]) -> int:
self.ans = 0
covered = {None}
def dfs(
cur: Optional[TreeNode],
parent: Optional[TreeNode]=None
) -> None:
if not cur:
return
dfs(cur.left, cur)
dfs(cur.right, cur)
if (
# if a node has no parent and it is not covered,
# we must place a camera here
(not parent and cur not in covered)
or
# if a node has children that are not covered by a camera,
# then we must place a camera here
(cur.left not in covered or cur.right not in covered)
):
self.ans += 1
covered.update({cur, parent, cur.left, cur.right})
dfs(root)
return self.ans
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