[LeetCode] 310. Minimum Height Trees

题目链接: https://leetcode.com/problems/minimum-height-trees/description/

Description

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.

解题思路

基本思路为按轮次删除入度为 1 的顶点，不断往中心缩减，最后只可能剩下 1 个或者 2 个顶点，返回剩下的顶点序号组成的数组即为目标解。

先将边数组 edges 转换为邻接表形式，并统计每个顶点的度。每轮要删除的顶点用队列保存，对每一个要删除的顶点，根据邻接表将其邻接顶点的度减 1，并判断减后度的值，若为 1 则入队。要注意一点，每轮开始前要统计一下该轮需要删除的顶点数，即先保存当前队列大小，然后该轮只 pop 上一轮 push 进队列的顶点，不可 pop 该轮新 push 进队列的顶点，否则就不是一轮一轮删除顶点，不能保证对称的往中间缩减。当只剩小于等于 2 个顶点时，终止删除过程，返回剩余顶点组成的数组。

Code

class Solution {public:    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {        vector<unordered_set<int>> adj(n);        vector<int> degree(n);        for (auto & e: edges) {            adj[e.first].insert(e.second);            adj[e.second].insert(e.first);            degree[e.first]++;            degree[e.second]++;        }        queue<int> todo;        for (int i = 0; i < n; i++) {            if (degree[i] <= 1) todo.push(i);        }        while (n > 2) {            int m = todo.size();            while (m-- > 0) {                int index = todo.front();                todo.pop();                n--;                for (int x: adj[index]) {                    if (--degree[x] == 1) todo.push(x);                }            }        }        vector<int> ret;        while (!todo.empty()) {            ret.push_back(todo.front());            todo.pop();        }        return ret;    }};