310. Minimum Height Trees

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 containsnnodes which are labeled from0ton - 1. You will be given the numbernand a list of undirectededges(each edge is a pair of labels).

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

Example 1:

Givenn = 4,edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3

return[1]

Example 2:

Givenn = 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 thedefinition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected byexactlyone 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.

S1: bfs, topological sort O(n)

leaf节点只有1个degree(parent), 其他节点至少有两个degree(parent + child)

类似topological sort从后往前，同时从所有leaf节点开始bfs往里找root，每次前进一步:删除所有leaf节点并更新他们的neighbors得到新一组leaf，直到最终剩余1-2个节点即为所求拥有最短路径的root。

    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        if (n == 1) return vector<int>{0}; //没有边的edge case
        vector<int> result;
        vector<unordered_set<int>> neighbors(n);
        vector<int> leafs; //类似queue的作用
        for (auto p : edges) {
            neighbors[p.first].insert(p.second);
            neighbors[p.second].insert(p.first);
        }
        for (int i = 0; i < n; ++i) {
            if (neighbors[i].size() == 1) leafs.push_back(i);
        }
        while (n > 2) {  //最多有两个min root
            vector<int> nextLeafs;
            for (int i : leafs) {
                for (int j : neighbors[i]) {
                    neighbors[j].erase(i);
                    if (neighbors[j].size() == 1) nextLeafs.push_back(j);
                }
                n--;
            }
            leafs = nextLeafs;
        }
        return leafs;
    }S2: dfs O(n^2)

计算每个节点的height

//超时
class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        vector<int> result;
        heights.resize(n);
        neighbors.resize(n);
        for (auto p : edges) {
            neighbors[p.first].push_back(p.second);
            neighbors[p.second].push_back(p.first);
        }
        minHeight = INT_MAX;
        for (int i = 0; i < n; ++i) {
            heights[i] = dfs(i, -1);
            minHeight = min(minHeight, heights[i]);
        }
        for (int i = 0; i < n; ++i) {
            if (heights[i] == minHeight) result.push_back(i);
        }
        return result;
    }

private:
    vector<int> heights;
    vector<vector<int>> neighbors;
    int minHeight;
    int dfs(int cur, int pre) {
        int h = 1;
        for (int i : neighbors[cur]) {
            if (i != pre) h = max(h, 1 + dfs(i, cur));
        }
        return h;
    }
};