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The greedy DFS solution is based on a bottom-up approach. We classify each node into one of three states: covered by a camera, has a camera, or needs a camera. A node needs a camera if its children are not covered. We use a post-order DFS strategy to ensure all children are addressed before deciding the current node's state. The optimal way to cover a subtree is to place a camera on nodes whose children are not covered.
Time Complexity: O(n), where n is the number of nodes in the tree, due to the DFS traversal of every node.
Space Complexity: O(h), where h is the height of the tree, due to the recursion stack space.
1public class TreeNode {
2 public int val;
3 public TreeNode left;
4 public TreeNode right;
5 public TreeNode(int value = 0) {
6 val = value;
7 }
8}
9
10public class Solution {
11 private int minCameras = 0;
12
13 public int MinCameraCover(TreeNode root) {
14 if (DFS(root) == 0) minCameras++;
15 return minCameras;
16 }
17
18 private int DFS(TreeNode node) {
19 if (node == null) return 2;
20 int left = DFS(node.left);
21 int right = DFS(node.right);
22 if (left == 0 || right == 0) {
23 minCameras++;
24 return 1;
25 }
26 return (left == 1 || right == 1) ? 2 : 0;
27 }
28}The C# solution defines a `TreeNode` class and uses a recursive DFS method (`DFS`). If a node's children need coverage, the node has a camera placed, updating `minCameras`.
This approach leverages dynamic programming through a recursive function with memoization to reduce redundant calculations. Each node can have three states represented in a tuple: it has a camera, it's covered but doesn't have a camera, or it's not covered. Using this state information, the solution calculates camera placements strategically.
Time Complexity: O(n), where n is the number of nodes.
Space Complexity: O(n), due to memoization storage.
1from functools import lru_cache
2
3class TreeNode:
4 def
This solution uses dynamic programming with memoization to avoid redundant recursive calls. The function `dfs` returns a tuple of three states for each node: the cost of having a camera, being covered without a camera, and not being covered.
The first element, `dp0`, calculates cost when a camera is placed at the current node. The second, `dp1`, calculates when the node is covered without having a camera, and the third, `dp2`, calculates when the node itself isn't covered.