[Swift]LeetCode968.监控二叉树 | Binary Tree Cameras

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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:

1. The number of nodes in the given tree will be in the range [1, 1000].
2. 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。

节点上的每个摄影头都可以监视其父对象、自身及其直接子对象。

计算监控树的所有节点所需的最小摄像头数量。

 

示例 1：

输入：[0,0,null,0,0]
输出：1
解释：如图所示，一台摄像头足以监控所有节点。

示例 2：

输入：[0,0,null,0,null,0,null,null,0]
输出：2
解释：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。

提示：

1. 给定树的节点数的范围是 [1, 1000]。
2. 每个节点的值都是 0。

---

 52ms

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func minCameraCover(_ root: TreeNode?) -> Int {
        var ans:[Int] = mh(root)
        return min(ans[1], ans[2])
    }
    
    func mh(_ ro:TreeNode?)-> [Int]
    {
        if ro == nil
        {
            var ans:[Int] = [Int](repeating:0,count:3)
            ans[1] = 1000000
            return ans
        }
        var l:[Int] = mh(ro!.left)
        var r:[Int] = mh(ro!.right)
        var ans:[Int] = [Int](repeating:0,count:3)
        ans[0] = l[2] + r[2];
        ans[1] = 1000000;
        ans[2] = 1000000;
        for i in 0..<3
        {
            for j in 0..<3
            {
                ans[1] = min(ans[1], l[i]+r[j]+1)
            }
            if i==0 {continue}
            ans[2] = min(ans[2], l[1]+r[i])
            ans[2] = min(ans[2], l[i]+r[1])
        }
        return ans
    }
}