968

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/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// 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;
    }
};