310. Minimum Height Trees

• Difficulty: Medium

• Topics: Breadth-first Search, Graph

• Similar Questions:

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Problem:

For an 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.

Solutions:

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<vector<int>>& edges) {
        if (n == 1) return {0}; // this check is important
        unordered_map<int, vector<int>> neighbors;
        unordered_map<int, int> degree;

        for (auto& edge : edges) {
            int node1 = edge[0];
            int node2 = edge[1];

            neighbors[node1].push_back(node2);
            neighbors[node2].push_back(node1);
            ++degree[node1];
            ++degree[node2];
        }

        vector<int> ret;
        queue<int> q;


        for (auto it = degree.begin(); it != degree.end(); ++it) {
            if (it->second == 1) {
                q.push(it->first);
            }
        }


        while(!q.empty()) {
            int size = q.size();
            ret.clear();
            for (int i = 0; i < size; ++i) {
                int node = q.front(); q.pop();
                //cout << node << endl;
                ret.push_back(node);
                for (auto& neighbor : neighbors[node]) {
                    if (--degree[neighbor] == 1) {
                        q.push(neighbor);
                    }
                }
            }
        }

        return ret;
    }
};