310-Minimum Height Trees

類別：graph

難度：medium

題目描述

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]




算法分析

（1）要使得樹的高度最小，只需要每一次的時候將所有的當前的度爲1的節點作爲葉子節點，同時該節點的度數置爲-1，剩餘節點數字減去1，並度爲1的所有葉子節點放在同一層，然後將這些節點刪掉並且將它們相鄰節點的度減去1，

（2）同樣的，在刪掉一層也葉子節點以後，再一次找當前的葉子節點，以此類推

（3）直到剩餘一個節點的時候，停止

（4）判斷還有多少節點的度>=0，即爲所求的根節點集合

代碼實現

class Solution {
public: 
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
    	vector<int> ans;
    	int edgeNum = edges.size();
    	vector< unordered_set<int>> adj(n);
    	for (int i = 0; i < edgeNum; ++i) {
    		adj[edges[i].first].insert(edges[i].second);
    		adj[edges[i].second].insert(edges[i].first);
		}
		vector<int> degree(n, 0);
		for (int i = 0; i < n; ++i) {
			degree[i] = adj[i].size();
		}
		int remainNum = n;
		while (remainNum > 2) {
            vector<int> delNodes;
			for (int i = 0; i < n; ++i) {
				if (degree[i] == 1) {
					remainNum--;
					delNodes.push_back(i);
					degree[i] = -1;
				}
			}
			for (int i = 0; i < delNodes.size(); ++i) {
                // 注意：unordered_set中的元素不能夠用訪問數組的方式進行訪問
				unordered_set<int> neighbors = adj[delNodes[i]]; 
				for (auto it = neighbors.begin(); it != neighbors.end(); ++it) {
					degree[*it]--;
				}
//				for (auto neighbor : adj[delNodes[i]]) {
//                    degree[neighbor]--;
//               }
			}
		}
		for (int i = 0; i < n; ++i) {
			if (degree[i] >= 0) ans.push_back(i);
		}
		return ans;
    }
};