<p>A tree is an undirected graph in which any two vertices are connected by&nbsp;<i>exactly</i>&nbsp;one path. In other words, any connected graph without simple cycles is a tree.</p>

<p>Given a tree of <code>n</code> nodes&nbsp;labelled from <code>0</code> to <code>n - 1</code>, and an array of&nbsp;<code>n - 1</code>&nbsp;<code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an undirected edge between the two nodes&nbsp;<code>a<sub>i</sub></code> and&nbsp;<code>b<sub>i</sub></code> in the tree,&nbsp;you can choose any node of the tree as the root. When you select a node <code>x</code> as the root, the result tree has height <code>h</code>. Among all possible rooted trees, those with minimum height (i.e. <code>min(h)</code>)&nbsp; are called <strong>minimum height trees</strong> (MHTs).</p>

<p>Return <em>a list of all <strong>MHTs&#39;</strong> root labels</em>.&nbsp;You can return the answer in <strong>any order</strong>.</p>

<p>The <strong>height</strong> of a rooted tree is the number of edges on the longest downward path between the root and a leaf.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/01/e1.jpg" style="width: 800px; height: 213px;" />
<pre>
<strong>Input:</strong> n = 4, edges = [[1,0],[1,2],[1,3]]
<strong>Output:</strong> [1]
<strong>Explanation:</strong> As shown, the height of the tree is 1 when the root is the node with label 1 which is the only MHT.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/01/e2.jpg" style="width: 800px; height: 321px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[3,0],[3,1],[3,2],[3,4],[5,4]]
<strong>Output:</strong> [3,4]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>All the pairs <code>(a<sub>i</sub>, b<sub>i</sub>)</code> are distinct.</li>
	<li>The given input is <strong>guaranteed</strong> to be a tree and there will be <strong>no repeated</strong> edges.</li>
</ul>

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<vector<int>>& edges) {
        vector<vector<int>> adj(n);
        for (auto e: edges) {
            adj[e[0]].push_back(e[1]);
            adj[e[1]].push_back(e[0]);
        }
        // first find diameter
        int dis[n];
        auto bfs_with_root = [n,edges,&dis,adj](int root) {
            queue<pair<int,int>> pq;
            int visited[n];
            for(int i=0;i<n;i++) {
                visited[i] = 0;
            }
            pq.push({0,root});
            visited[root] = 1;
            int max_dis_index = root;
            while(!pq.empty()) {
                auto p = pq.front();
                pq.pop();
                int distance = p.first;
                int index = p.second;
                //cout << distance<<" "<< index << ", ";
                dis[index] = distance;
                if(distance > dis[max_dis_index]) {
                    max_dis_index = index;
                }
                for (auto v : adj[index]) {
                    if(!visited[v]) {
                        pq.push({distance+1,v});
                        visited[v] = 1;
                    }
                }
            }
            return max_dis_index;
        };
        int index1 = bfs_with_root(0);
        int index2 = bfs_with_root(index1);
        int diameter = dis[index2];
        int result1[n];
        for(int i=0;i<n;i++)
        result1[i] = dis[i];
        bfs_with_root(index2);
        vector<int> results;
        int t2 = (diameter+1)/2;
        for(int i=0;i<n;i++) {
            if(result1[i] <= t2 && dis[i] <= t2 && result1[i]+dis[i] == diameter) {
                results.push_back(i);
            }
        }
        return results; 
    }
};
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