Minimum Height Trees -- LeetCode

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]

思路：找到这棵树中所有度为1的节点，它们是最外层的节点。从树中移除这些节点，重复这个操作直到树中的节点数小于3。结果可能是1个节点或者两个节点，这取决于树中最长路径的长度（奇偶性）。时间复杂度O(N)。

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        if (n == 1) return vector<int>(1, 0);
        vector<unordered_set<int> > tree(n, unordered_set<int>());
        for (int i = 0; i < edges.size(); i++) {
            tree[edges[i].first].insert(edges[i].second);
            tree[edges[i].second].insert(edges[i].first);
        }
        vector<int> leaves;
        for (int i = 0; i < n; i++)
            if (tree[i].size() == 1) leaves.push_back(i);
        while (n > 2) {
            vector<int> newLeaves;
            for (int i = 0; i < leaves.size(); i++)
                for (auto j : tree[leaves[i]]) {
                    tree[j].erase(leaves[i]);
                    if (tree[j].size() == 1) newLeaves.push_back(j);
                }
            n -= leaves.size();
            leaves = newLeaves;
        }
        return leaves;
    }
};