968. Binary Tree Cameras

LeetCode hard original: C# #array #csharp #hard #leetcode #math #recursion #tree #two-pointers

선택한 UI 언어에 맞게 문제 텍스트를 러시아어에서 번역합니다. 코드는 변경하지 않습니다.

LeetCode Static

Вам дан корень бинарного дерева. Мы устанавливаем камеры на узлы дерева, где каждая камера на узле может наблюдать за своим родителем, собой и своими непосредственными детьми.

return минимальное количество камер, необходимых для наблюдения за всеми узлами дерева.

예제:

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.

C# 해법

매칭됨/원본

public class Solution {
    public int MinCameraCover(TreeNode root) {
        var ans = Solve(root);
        return Math.Min(ans[1], ans[2]);
    }
    private int[] Solve(TreeNode node) {
        if (node == null) {
            return new int[] { 0, 0, 99999 };
        }
        var L = Solve(node.left);
        var R = Solve(node.right);
        int mL12 = Math.Min(L[1], L[2]);
        int mR12 = Math.Min(R[1], R[2]);
        int d0 = L[1] + R[1];
        int d1 = Math.Min(L[2] + mR12, R[2] + mL12);
        int d2 = 1 + Math.Min(L[0], mL12) + Math.Min(R[0], mR12);
        return new int[] { d0, d1, d2 };
    }
}

C++ 해법

자동 초안, 제출 전 검토

#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
    public int MinCameraCover(TreeNode root) {
        var ans = Solve(root);
        return min(ans[1], ans[2]);
    }
    private vector<int>& Solve(TreeNode node) {
        if (node == null) {
            return new int[] { 0, 0, 99999 };
        }
        var L = Solve(node.left);
        var R = Solve(node.right);
        int mL12 = min(L[1], L[2]);
        int mR12 = min(R[1], R[2]);
        int d0 = L[1] + R[1];
        int d1 = min(L[2] + mR12, R[2] + mL12);
        int d2 = 1 + min(L[0], mL12) + min(R[0], mR12);
        return new int[] { d0, d1, d2 };
    }
}

Java 해법

매칭됨/원본

class Solution {
    public int minCameraCover(TreeNode root) {
        int[] ans = solve(root);
        return Math.min(ans[1], ans[2]);
    }

    public int[] solve(TreeNode node) {
        if (node == null)
            return new int[]{0, 0, 99999};

        int[] L = solve(node.left);
        int[] R = solve(node.right);
        int mL12 = Math.min(L[1], L[2]);
        int mR12 = Math.min(R[1], R[2]);

        int d0 = L[1] + R[1];
        int d1 = Math.min(L[2] + mR12, R[2] + mL12);
        int d2 = 1 + Math.min(L[0], mL12) + Math.min(R[0], mR12);
        return new int[]{d0, d1, d2};
    }
}

JavaScript 해법

매칭됨/원본

var minCameraCover = function(root) {
    const solve = (node) => {
        if (!node) return [0, 0, 99999]

        const L = solve(node.left)
        const R = solve(node.right)
        const mL12 = Math.min(L[1], L[2])
        const mR12 = Math.min(R[1], R[2])

        const d0 = L[1] + R[1]
        const d1 = Math.min(L[2] + mR12, R[2] + mL12)
        const d2 = 1 + Math.min(L[0], mL12) + Math.min(R[0], mR12)
        return [d0, d1, d2]
    }

    const ans = solve(root)
    return Math.min(ans[1], ans[2])
}

Python 해법

매칭됨/원본

class Solution:
    def minCameraCover(self, root: TreeNode) -> int:
        def solve(node):
            if not node:
                return 0, 0, float('inf')

            L = solve(node.left)
            R = solve(node.right)
            mL12 = min(L[1], L[2])
            mR12 = min(R[1], R[2])

            d0 = L[1] + R[1]
            d1 = min(L[2] + mR12, R[2] + mL12)
            d2 = 1 + min(L[0], mL12) + min(R[0], mR12)
            return d0, d1, d2

        return min(solve(root)[1:])

Vacancies for this task

활성 채용 with overlapping task tags are 표시됨.

전체 채용

아직 활성 채용이 없습니다.