310. Minimum Height Trees

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题目描述

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]

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.

题意分析

给定一个无向图，从中选取一个节点作为根节点, 看做一棵树，返回最小高度的根节点(所有)。

算法分析

当n=1, 树高只有1, 直接返回

当n=2，选任意的一个点都是一样的, 两个点都要返回

当n>2，符合条件的节点最多只有两个. 因为假设存在三个点，说明以这三个点为root的树高度是一样的，就是到达最远叶子节点的路径长度是一样的，现在我们逐步删除叶子节点，这样三个点为root的树高度应该逐步减1，最后如果高度都减为0，说明有三个根节点也就是三棵树，这三个节点是独立的，违背了树的定义。

所以我们从叶子节点出发, 做BFS, 遍历过的叶子节点删除掉, 将新成为叶子的节点加入队列中, 直到最后删剩的节点数少于2.

实现

class Solution {public:    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {        if (n == 1) return {0};        vector< unordered_set<int> > adjMatrix(n);        for (auto edge : edges) {       // add edges to matrix            adjMatrix[edge.first].insert(edge.second);            adjMatrix[edge.second].insert(edge.first);        }        queue<int> leaf;        for (int i = 0; i < n; i++) {            if (adjMatrix[i].size() == 1)                 leaf.push(i);        }        while(n > 2) {            int size = leaf.size();            n -= size;            for (int i = 0; i < size; ++i) {  // 要一次删去一层节点,与以往bfs的写法略有不同                int temp = leaf.front();                leaf.pop();                for (auto adjV : adjMatrix[temp]) {                    adjMatrix[adjV].erase(temp);                    if (adjMatrix[adjV].size() == 1)                         leaf.push(adjV);                }            }        }        vector<int> result;        while (!leaf.empty()) {            result.push_back(leaf.front());            leaf.pop();        }        return result;    }};

复杂度分析

时间复杂度 O(|V|+|E|)
空间复杂度 O(|V|+|E|)