Formatted question description: https://leetcode.ca/all/968.html

968. Binary Tree Cameras (Hard)

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.

Companies:
Facebook

Related Topics:
Dynamic Programming, Tree, Depth-first Search

Solution 1.

Assumption: we can alter the value of the tree node.

Use postorder traversal. Let val convey the state of the node:

  • 0 means uncovered.
  • 1 means covered
  • 2 means having camera

If node is NULL, we regard it as 1.

  • If either of my left/right child is uncovered, I have to put a camera.
  • Otherwise, if either of my left/right child has camera, I’m covered and skip me.
  • Otherwise (both children are covered), if I’m root, I have to put a camera.
  • Otherwise, skip me.
// OJ: https://leetcode.com/problems/binary-tree-cameras/

// Time: O(N)
// Space: O(logN)
class Solution {
private:
    TreeNode *R;
    int postorder(TreeNode* root) {
        if (!root) return 0;
        int ans = postorder(root->left) + postorder(root->right);
        int left = root->left ? root->left->val : 1;
        int right = root->right ? root->right->val : 1;
        if (left == 0 || right == 0) {
            root->val = 2;
            return ans + 1;
        } else if (left == 2 || right == 2) {
            root->val = 1;
            return ans;
        } else return root == R ? ans + 1 : ans;
    }
public:
    int minCameraCover(TreeNode* root) {
        R = root;
        return postorder(root);
    }
};

Solution 2.

If we can’t have the assumption in Solution 1, we can use the return value to return my state to my parent.

// OJ: https://leetcode.com/problems/binary-tree-cameras/

// Time: O(N)
// Space: O(logN)
class Solution {
private:
    int ans = 0;
    int postorder(TreeNode *root) {
        if (!root) return 1;
        int left = postorder(root->left);
        int right = postorder(root->right);
        if (left == 0 || right == 0) {
            ++ans;
            return 2;
        } else return left == 2 || right == 2 ? 1 : 0;
    }
public:
    int minCameraCover(TreeNode* root) {
        return postorder(root) == 0 ? ans + 1 : ans;
    }
};

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int minCameraCover(TreeNode root) {
        if (root == null)
            return 0;
        List<TreeNode> list = new ArrayList<TreeNode>();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            list.add(node);
            TreeNode left = node.left, right = node.right;
            if (left != null)
                queue.offer(left);
            if (right != null)
                queue.offer(right);
        }
        Map<TreeNode, int[]> map = new HashMap<TreeNode, int[]>();
        for (int i = list.size() - 1; i >= 0; i--) {
            TreeNode node = list.get(i);
            if (node.left == null && node.right == null)
                map.put(node, new int[]{0, Integer.MAX_VALUE / 10, 1});
            else if (node.left == null) {
                int[] rightArray = map.get(node.right);
                int[] array = new int[3];
                array[0] = rightArray[1];
                array[1] = rightArray[2];
                array[2] = 1 + Math.min(rightArray[0], Math.min(rightArray[1], rightArray[2]));
                map.put(node, array);
            } else if (node.right == null) {
                int[] leftArray = map.get(node.left);
                int[] array = new int[3];
                array[0] = leftArray[1];
                array[1] = leftArray[2];
                array[2] = 1 + Math.min(leftArray[0], Math.min(leftArray[1], leftArray[2]));
                map.put(node, array);
            } else {
                int[] leftArray = map.get(node.left);
                int[] rightArray = map.get(node.right);
                int minLeft = Math.min(leftArray[1], leftArray[2]);
                int minRight = Math.min(rightArray[1], rightArray[2]);
                int[] array = new int[3];
                array[0] = leftArray[1] + rightArray[1];
                array[1] = Math.min(leftArray[2] + minRight, rightArray[2] + minLeft);
                array[2] = 1 + Math.min(leftArray[0], minLeft) + Math.min(rightArray[0], minRight);
                map.put(node, array);
            }
        }
        int[] rootArray = map.get(root);
        return Math.min(rootArray[1], rootArray[2]);
    }
}

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