Given a binary tree, we install cameras on the nodes of the tree.
Each camera at a node can monitor its parent, itself, and its immediate children.
Calculate the minimum number of cameras needed to monitor all nodes of the tree.
Example 1:
Input: [0,0,null,0,0]
Output: 1
Explanation: One camera is enough to monitor all nodes if placed as shown.
Example 2:
Input: [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.
Note:
- The number of nodes in the given tree will be in the range
[1, 1000]. - Every node has value 0.
给定一个二叉树，我们在树的节点上安装摄像头。
节点上的每个摄像头都可以监视其父级、自身及其直接子级。
计算监控树的所有节点所需的最少摄像头数量。
例1：
输入：[0,0,null,0,0]
输出：1
说明：如果按图中所示放置，一个摄像头足以监视所有节点。
例2：
输入：[0,0,null,0,null,0,null,null,0]
输出：2
说明：至少需要两个摄像头来监控树的所有节点。上图显示了相机放置的有效配置之一。
注：
- 给定树中的节点数将在范围[1000]内。
- 每个节点都有值0。
52ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil 10 * self.right = nil 11 * } 12 * } 13 */ 14 class Solution { 15 func minCameraCover(_ root: TreeNode?) -> Int { 16 var ans:[Int] = mh(root) 17 return min(ans[1], ans[2]) 18 } 19 20 func mh(_ ro:TreeNode?)-> [Int] 21 { 22 if ro == nil 23 { 24 var ans:[Int] = [Int](repeating:0,count:3) 25 ans[1] = 1000000 26 return ans 27 } 28 var l:[Int] = mh(ro!.left) 29 var r:[Int] = mh(ro!.right) 30 var ans:[Int] = [Int](repeating:0,count:3) 31 ans[0] = l[2] + r[2]; 32 ans[1] = 1000000; 33 ans[2] = 1000000; 34 for i in 0..<3 35 { 36 for j in 0..<3 37 { 38 ans[1] = min(ans[1], l[i]+r[j]+1) 39 } 40 if i==0 {continue} 41 ans[2] = min(ans[2], l[1]+r[i]) 42 ans[2] = min(ans[2], l[i]+r[1]) 43 } 44 return ans 45 } 46 }