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968. Binary Tree Cameras

Description

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

Solutions

Solution 1

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# 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:
        def dfs(root):
            if root is None:
                return inf, 0, 0
            la, lb, lc = dfs(root.left)
            ra, rb, rc = dfs(root.right)
            a = min(la, lb, lc) + min(ra, rb, rc) + 1
            b = min(la + rb, lb + ra, la + ra)
            c = lb + rb
            return a, b, c

        a, b, _ = dfs(root)
        return min(a, b)
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/**
 * 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 {
    public int minCameraCover(TreeNode root) {
        int[] ans = dfs(root);
        return Math.min(ans[0], ans[1]);
    }

    private int[] dfs(TreeNode root) {
        if (root == null) {
            return new int[] {1 << 29, 0, 0};
        }
        var l = dfs(root.left);
        var r = dfs(root.right);
        int a = 1 + Math.min(Math.min(l[0], l[1]), l[2]) + Math.min(Math.min(r[0], r[1]), r[2]);
        int b = Math.min(Math.min(l[0] + r[1], l[1] + r[0]), l[0] + r[0]);
        int c = l[1] + r[1];
        return new int[] {a, b, c};
    }
}
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
struct Status {
    int a, b, c;
};

class Solution {
public:
    int minCameraCover(TreeNode* root) {
        auto [a, b, _] = dfs(root);
        return min(a, b);
    }

    Status dfs(TreeNode* root) {
        if (!root) {
            return {1 << 29, 0, 0};
        }
        auto [la, lb, lc] = dfs(root->left);
        auto [ra, rb, rc] = dfs(root->right);
        int a = 1 + min({la, lb, lc}) + min({ra, rb, rc});
        int b = min({la + ra, la + rb, lb + ra});
        int c = lb + rb;
        return {a, b, c};
    };
};
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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func minCameraCover(root *TreeNode) int {
    var dfs func(*TreeNode) (int, int, int)
    dfs =<