0968. Binary Tree Cameras (H)

题目

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.

题意

在二叉树中选取任意个结点放上一个照相机，每个照相机能覆盖当前结点、当前结点的父结点、当前结点的直接子结点。问最少需要放几个照相机能够覆盖所有结点。

思路

贪心。从最底下的结点开始往上，对于当前结点有三种情况需要放置照相机：(1) 当前结点的左子结点未被覆盖；(2) 当前结点的右子结点未被覆盖；(3) 当前结点未被覆盖，且无父结点。其他情况无需放置，最大化每个照相机的覆盖范围。

代码实现

Java

class Solution {
    private int count;
    private Set<TreeNode> set;
    public int minCameraCover(TreeNode root) {
        count = 0;
        set = new HashSet<>();
        set.add(null);
        dfs(root, null);
        return count;
    }
    private void dfs(TreeNode root, TreeNode parent) {
        if (root == null) return;
        dfs(root.left, root);
        dfs(root.right, root);
        if (parent == null && !set.contains(root) || !set.contains(root.left) || !set.contains(root.right)) {
            set.add(root);
            set.add(root.left);
            set.add(root.right);
            set.add(parent);
            count++;
        }
    }
}