[Leetcode] 310. Minimum Height Trees 解题报告_return a list of their root labels.-优快云博客

本文介绍了一种算法，用于找出给定树形图中所有最小高度的根节点。通过使用广度优先搜索(BFS)策略去除度数为1的节点，最终确定树的根节点。文章提供了详细的代码实现。

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题目：

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:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

return [3, 4]

Note:

(1) 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.”

(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

思路：

刚开始觉得是一道拓扑排序的题目：每次都去掉顶点度数为1的点；最后当图中只剩下小于等于2个结点时就可以返回结果了。后来发现实现的时候用BFS最方便：首先将度数为1的顶点收集起来，然后逐个从图中将这些点删除；当然在删除的过程中，又会出现新的度数为1的顶点，所以我们需要将这些点收集起来并在下一轮中继续删除。这样迭代直到图中顶点的个数小于等于2，就可以返回结果了。

代码：

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        unordered_map<int, unordered_set<int>> hash;    // build and initiate the graph using the hash table
        for(auto val : edges) {
            hash[val.first].insert(val.second);
            hash[val.second].insert(val.first);
        }
        queue<int> q;                                   // define and initialize the queue for BFS
        for(int i = 0; i < n; ++i) {
            if(hash[i].size() == 1)
                q.push(i);
        }
        while(hash.size() > 2) {                        // Update the hash by erasing the nodes with 1 degree
            queue<int> temp;
            while(!q.empty()) {
                int node1 = q.front();
                q.pop();
                int node2 = *(hash[node1].begin());
                hash[node2].erase(node1);
                if(hash[node2].size() == 1) {
                    temp.push(node2);
                }
                hash.erase(node1);
            }
            q = temp;
        }
        vector<int> ret;                                // construct the return value
        for(auto val : hash) {
            ret.push_back(val.first);
        }
        return ret;
    }
};