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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.
1class TreeNode {
2 constructor(val = 0, left = null, right = null) {
3 this.val = val;
4 this.left = left;
5 this.right = right;
6 }
7}
8
9var minCameraCover = function(root) {
10 let minCameras = 0;
11
12 const dfs = function(node) {
13 if (!node) return 2;
14 let left = dfs(node.left);
15 let right = dfs(node.right);
16 if (left === 0 || right === 0) {
17 minCameras++;
18 return 1;
19 }
20 return (left === 1 || right === 1) ? 2 : 0;
21 };
22
23 if (dfs(root) === 0) minCameras++;
24 return minCameras;
25};The JavaScript solution uses a recursive function `dfs` within `minCameraCover`. A camera is added whenever a node's children are not covered. The `minCameras` count is incremented when necessary based on the state returned from the children.
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.