LeetCode #310 - Minimum Height Trees

最新推荐文章于 2023-02-25 22:41:55 发布

LawFile

最新推荐文章于 2023-02-25 22:41:55 发布

阅读量128

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/81199824

LeetCode 专栏收录该内容

625 篇文章

订阅专栏

本文介绍了一种高效算法，用于找到给定无向图中能够成为最小高度树的所有根节点。通过构建邻接表并计算节点度数，逐步去除叶节点直至找到符合条件的根节点。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges(each edge is a pair of labels).

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Example 1 :

Input: n = 4, edges = [[1, 0], [1, 2], [1, 3]]        0
        |
        1
       / \
      2   3 

Output: [1]

Example 2 :

Input: n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]
     0  1  2
      \ | /
        3
        |
        4
        |
        5 
Output: [3, 4]

Note:

According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”
The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

给定一个树，可以选取任意节点作为根节点，求选取哪些节点为根节点时，树的高度最小。逐个选取根节点计算树高度的时间复杂度过高，可以先构建邻接表并计算每个节点的度，度为1说明是叶节点。那么从叶节点出发，向邻接点遍历，每次遍历对应的节点度数减一，如果度数变为1又可以加入队列，直到所有节点都访问到，那么最后一次遍历的一组节点就是所求。另外由于我们讨论的对象是树，也就是说每个节点最多有两个邻接点，且不存在环，那么最终满足要求的根节点只可能是1个或者2个，以此作为循环结束的依据。

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        if(n==1) return vector<int>(1,0);
        vector<vector<int>> graph(n);
        vector<int> degree(n,0);
        for(int i=0;i<edges.size();i++)
        {
            degree[edges[i].first]++;
            degree[edges[i].second]++;
            graph[edges[i].first].push_back(edges[i].second);
            graph[edges[i].second].push_back(edges[i].first);
        }
        
        queue<int> q;
        for(int i=0;i<n;i++) if(degree[i]==1) q.push(i);
        while(n>2)
        {
            int count=q.size();
            n-=count;
            for(int k=0;k<count;k++)
            {
                int i=q.front();
                q.pop();
                for(int j=0;j<graph[i].size();j++)
                {
                    degree[graph[i][j]]--;
                    if(degree[graph[i][j]]==1) q.push(graph[i][j]);
                }
            }
        }
        
        vector<int> result;
        while(!q.empty())
        {
            result.push_back(q.front());
            q.pop();
        }
        return result;
    }
};