【Leetcode】310. Minimum Height Trees

Description:

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:

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

    0    |    1   / \  2   3

return [1]

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

 0  1  2  \ | /    3    |    4    |    5

return [3, 4]

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思路：

根据题目要求，在给定的无向图中找到所有的最小高度树，并返回树的根节点。最初的基本思想，以任意一个点作为根节点，然后找到其最大深度，这个深度则是树的高度。对所有的点都进行这样的操作，找出每个点作为根节点的树的高度，相比较则可以找出最小高度和最小高度树，最后统计其数量。
另外，基于BFS算法，可以通过找出所有入度（或出度，由于本题是无向图，所以入度与出度的值一致）为1的点，逐步对这些点进行遍历删除，那么最后剩下的点则是最小高度树的根节点。
下面是使用了C++的实现方法：

class Solution { public:  vector<int> findMinHeightTrees(int n, vector<pair<int, int> >& edges) {    vector<vector<int> > G(n);    //构造图，使得每个点存储其所相连的所有点    for(int i=0; i<edges.size(); i++){        G[edges[i].first].insert(edges[i].second);        G[edges[i].second].insert(edges[i].first);    }    //当只有一个节点时    vector<int> cur;    if(n==1){        cur.push_back(0);        return cur;    }    //存储入度出度都为1的点    for(int i=0; i<G.size(); i++){        if (G[i].size() == 1) {            cur.push_back(i);        }    }    //对图进行BFS操作    while(true){        vector<int> next_visit;        for (int i=0;i<cur.size();i++) {            int node = cur[i];            for (int j=0;j<G[node].size();j++) {                int neighbor = G[node][j];                vector<int>::iterator iter = find(G[neighbor].begin(),G[neighbor].end(),node);//利用迭代器存储G[neighbor]中值与node相同的元素                G[neighbor].erase(iter);                if (G[neighbor].size() == 1) next_visit.push_back(neighbor);            }        }        if (next_visit.empty()) return cur;        cur = next_visit;         }    }};