<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 MyGraphNode {
public:
int val;
vector<int> neighbours;
    MyGraphNode(int val) {
        this->val = val;
    }
};

class Solution {
unordered_map<int, MyGraphNode*> graph;
unordered_map<int, bool> visited;

private:
    int bfsFurthestNode(int n) {
        visited.clear();
        queue<int> fifo;
        fifo.push(n);
        int answer = n;
        while (fifo.size()) {
            int current = fifo.front();
            answer = current;
            fifo.pop();
            visited[current] = true;
            for (int neighbour : graph[current]->neighbours) {
                if (visited[neighbour]) continue;
                fifo.push(neighbour);
            }
        }
        return answer;
    }

    void dfsGetPath(int current, int goal, vector<int>& stack) {
        if (stack.size() && stack.back() == goal) return; 
        if (visited[current]) return;
        visited[current] = true;
        stack.push_back(current);
        if (current == goal) return;
            for (int neighbour : graph[current]->neighbours) {
            if (visited[neighbour]) continue;
            dfsGetPath(neighbour, goal, stack);
            if (stack.size() && stack.back() == goal) return; 
        }
        stack.pop_back();
    }

public:
    vector<int> findMinHeightTrees(int n, vector<vector<int>>& edges) {
        graph.clear();
        for (int i = 0; i < n; i++) {
            graph[i] = new MyGraphNode(i);
        }
        for (vector<int>& edge : edges) {
            graph[edge[1]]->neighbours.push_back(edge[0]);
            graph[edge[0]]->neighbours.push_back(edge[1]);
        }
        int B = bfsFurthestNode(0);
        // cout << "furthest from 0: " << B << endl;
        int C = bfsFurthestNode(B);
        // cout << "furthest from above: " << C << endl;
        visited.clear();
        vector<int> diameter = {};
        dfsGetPath(B, C, diameter);
        // for (int n : diameter) {
        //     cout << n << " -> ";
        // }
        // cout << endl;
        if (diameter.size() & 1) return {diameter[diameter.size()/2]};
        else return {
            diameter[diameter.size()/2],
            diameter[(diameter.size()/2)-1],
        };
    }
};
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