<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:
    void DFS(int src, int parent, vector<bool>& visited, vector<vector<int> >& tree, vector<pair<int, int> >& dp, vector<int>& par){
        visited[src] = true;
        par[src] = parent;
        for(int i=0;i<tree[src].size();i++){
            int neighbor = tree[src][i];
            if(neighbor == parent)
                continue;
            if(!visited[neighbor])
                DFS(neighbor, src, visited, tree, dp, par);
            if(dp[src].first < 1 + dp[neighbor].first){
                dp[src].first = 1 + dp[neighbor].first;
                dp[src].second = dp[neighbor].second;
            }
        }
        if(dp[src].first == -1){
            dp[src].first = 0;
            dp[src].second = src;
        }
    }
    vector<int> findMinHeightTrees(int n, vector<vector<int>>& edges) {
        vector<vector<int> > tree(n);
        for(int i=0;i<edges.size();i++){
            int a = edges[i][0];
            int b = edges[i][1];
            tree[a].push_back(b);
            tree[b].push_back(a);
        }
        vector<bool> visited(n, false);
        vector<int> parent(n, -1);
        // depth, lowest vertex
        vector<pair<int, int> > dp(n, make_pair(-1, 0));
        DFS(0, -1, visited, tree, dp, parent);
        int leaf = dp[0].second;
        // cout << leaf << endl;
        vector<pair<int, int> > dp2(n, make_pair(-1, 0));
        vector<bool> visited2(n, false);
        vector<int> parent2(n, -1);
        DFS(leaf, -1, visited2, tree, dp2, parent2);
        int other_leaf = dp2[leaf].second;
        // cout << other_leaf << endl;
        vector<int> diameter;
        diameter.push_back(other_leaf);
        int p = parent2[other_leaf];
        while(p != -1){
            diameter.push_back(p);
            p = parent2[p];
        }
        int size = diameter.size();
        vector<int> ans;
        if(size % 2 == 0){
            ans.push_back(diameter[size/2]);
            ans.push_back(diameter[size/2 - 1]);
        }
        else
            ans.push_back(diameter[size/2]);
        return ans;
    }
};
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