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:
The number of nodes in the given tree will be in the range [1, 1000].
Every node has value 0.
Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int NOT_MONITORED = 0;
private int MONITORED_NOCAM = 1;
private int MONITORED_CAM = 2;
public int minCameraCover(TreeNode root) {
if (root == null) return 0;
int[] count = new int[]{0};
int top = helper(root, count);
return count[0] + (top == NOT_MONITORED ? 1 : 0);
}
private int helper(TreeNode root, int[] count) {
if (root == null) {
return MONITORED_NOCAM;
}
int left = helper(root.left, count);
int right = helper(root.right, count);
if (left == NOT_MONITORED || right == NOT_MONITORED) {
count[0] ++;
return MONITORED_CAM;
} else if (left == MONITORED_NOCAM && right == MONITORED_NOCAM) {
return NOT_MONITORED;
} else {
return MONITORED_NOCAM;
}
}
}