LeetCode

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]]

        0        |        1       / \      2   3

return [1]

Example 2:

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

     0  1  2      \ | /        3        |        4        |        5

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.

给定一个拥有树性质的无向图，图的每一个节点都可以视为一棵树的根节点。在所有可能的树中，找出高度最小的树，并返回他们的树根。

这个思路实际上是一个 BFS 思路。和常见的从根节点进行 BFS 不同，这里从叶子节点开始进行 BFS。

所有入度（即相连边数）为 1 的节点即是叶子节点。找高度最小的节点，即找离所有叶子节点最远的节点，也即找最中心的节点

找最中心的节点的思路很简单：

• 每次去掉当前图的所有叶子节点，重复此操作直到只剩下最后的根。

根据这个思路，只能有一个或者两个最小高度树树根。证明省略。

class Solution {public:    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {        vector<int> ans;        if (n <= 1) {            ans.push_back(0);            return ans;        }                vector<unordered_set<int> > edg(n);        for (auto x: edges) {            edg[x.first].insert(x.second);            edg[x.second].insert(x.first);        }        vector<int> degree(n, 0);        for (int i = 0; i < n; ++i)            degree[i] = edg[i].size();                for (int cnt = n; cnt > 2; ) {            vector<int> cur;            for (int i = 0; i < n; ++i) {   //找到叶子结点                if (degree[i] == 1) {                    cur.push_back(i);                    cnt--;                    degree[i] = -1;                }            }            for (int i = 0; i < cur.size(); ++i) {  //删除这些点和对应的边                for (auto x: edg[cur[i]]) {                    degree[x]--;                }            }        }                for (int i = 0; i < n; ++i) {//得出答案            if (degree[i] >= 0) ans.push_back(i);        }        return ans;    }};