LeetCode in Kotlin
968. Binary Tree Cameras
Hard
You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.
Return the minimum number of cameras needed to monitor all nodes of the tree.
Example 1:
Input: root = [0,0,null,0,0]
Output: 1
Explanation: One camera is enough to monitor all nodes if placed as shown.
Example 2:
Input: root = [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.
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. Node.val == 0
Solution
import com_github_leetcode.TreeNode
/*
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Solution {
private var cameras = 0
fun minCameraCover(root: TreeNode?): Int {
cameras = 0
if (minCameras(root) == -1) {
// root needs a camera
cameras++
}
return cameras
}
// States =>
// -1 : Node needs a camera
// 0 : Node has a camera placed
// 1 : Node is covered somehow
private fun minCameras(root: TreeNode?): Int {
if (root == null) {
return 1
}
val leftChildState = minCameras(root.left)
val rightChildState = minCameras(root.right)
// One of the two or both children need a camera
if (leftChildState == -1 || rightChildState == -1) {
// installed
cameras++
return 0
}
// One of the two or both children have a camera placed
return if (leftChildState == 0 || rightChildState == 0) {
// gets covered by the children
1
} else {
-1
}
// needs a camera
}
}