310. Minimum Height Trees-优快云博客

本文介绍了一种算法，用于找出特定类型的图（具有树特征的无向图）中的最小高度树。通过逐步移除叶子节点的方法，最终确定图中作为根节点能够构成最小高度树的所有节点。

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 exactlyone 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.

Credits:

Special thanks to @dietpepsi for adding this problem and creating all test cases.

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        vector<int> res;
        if(n==1)
        {
            res.push_back(0);
            return res;
        }
        
        unordered_map<int ,unordered_set<int> > adj;
        for(auto &entry : edges)
        {
            adj[entry.first].insert(entry.second);
            adj[entry.second].insert(entry.first);
        }
        
        vector<int> leaves;
        for(auto &entry : adj)
        {
            if(adj[entry.first].size()==1)
            {
                leaves.push_back(entry.first);
            }
        }
        
        while(n>2)
        {
            n -= leaves.size();
            vector<int> nextLeaves;
            
            for(int idx=0; idx<leaves.size(); idx++)
            {
                int nextNode = *(adj[leaves[idx]].begin());
                adj[leaves[idx]].erase(nextNode);
                adj[nextNode].erase(leaves[idx]);
                
                if(adj[nextNode].size()==1)
                {
                    nextLeaves.push_back(nextNode);
                }
            }
            
            leaves = nextLeaves;
            
        }
        return leaves;
    }
};