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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:

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;
}
};

• /**
* 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.offer(root);
while (!queue.isEmpty()) {
TreeNode node = queue.poll();
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]);
}
}

############

/**
* 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 ans;

public int minCameraCover(TreeNode root) {
ans = 0;
return (dfs(root) == 0) ? ans + 1 : ans;
}

private int dfs(TreeNode root) {
if (root == null) {
return 2;
}
int left = dfs(root.left);
int right = dfs(root.right);
if (left == 0 || right == 0) {
++ans;
return 1;
}
if (left == 1 || right == 1) {
return 2;
}
return 0;
}
}

• // 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);
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def minCameraCover(self, root: TreeNode) -> int:
def dfs(root):
nonlocal ans
if root is None:
return 2
left, right = dfs(root.left), dfs(root.right)
if left == 0 or right == 0:
ans += 1
return 1
return 2 if left == 1 or right == 1 else 0

ans = 0
return (dfs(root) == 0) + ans

############

# 968. Binary Tree Cameras
# https://leetcode.com/problems/binary-tree-cameras/

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def minCameraCover(self, root: Optional[TreeNode]) -> int:
res = 0

def go(node):
nonlocal res

if not node: return 2

left, right = go(node.left), go(node.right)

if left == 0 or right == 0:
res += 1
return 1

return 2 if left == 1 or right == 1 else 0

return (go(root) == 0) + res


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func minCameraCover(root *TreeNode) int {
ans := 0
var dfs func(root *TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 2
}
left, right := dfs(root.Left), dfs(root.Right)
if left == 0 || right == 0 {
ans++
return 1
}
if left == 1 || right == 1 {
return 2
}
return 0
}
if dfs(root) == 0 {
return ans + 1
}
return ans
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function minCameraCover(root: TreeNode | null): number {
const dfs = (root: TreeNode | null): number[] => {
if (!root) {
return [1 << 29, 0, 0];
}
const [la, lb, lc] = dfs(root.left);
const [ra, rb, rc] = dfs(root.right);
const a = 1 + Math.min(la, lb, lc) + Math.min(ra, rb, rc);
const b = Math.min(la + ra, la + rb, lb + ra);
const c = lb + rb;
return [a, b, c];
};
const [a, b, _] = dfs(root);
return Math.min(a, b);
}