<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) {
        if (n == 1) return {0};  // Special case for a single node
        
        // Build the adjacency list
        vector<vector<int>> adj(n);
        for (const auto& edge : edges) {
            adj[edge[0]].push_back(edge[1]);
            adj[edge[1]].push_back(edge[0]);
        }

        // Find the farthest node from an arbitrary node (let's use 0)
        int nodeA = findFarthestNode(0, adj).first;
        
        // Find the farthest node from nodeA, which gives us the other endpoint of the diameter
        pair<int, int> second_search = findFarthestNode(nodeA, adj);
        int nodeB = second_search.first;
        int diameter_length = second_search.second;
        
        // Get the path length and diameter
        vector<int> distA = bfs(nodeA, adj);
        vector<int> distB = bfs(nodeB, adj);
        
        vector<int> result;
        int half = diameter_length / 2;
        for (int i = 0; i < n; ++i) {
            if (distA[i] == half && distB[i] == diameter_length - half) {
                result.push_back(i);
            } else if (distB[i] == half && distA[i] == diameter_length - half) {
                result.push_back(i);
            }
        }

        return result;
    }

private:
    // BFS to find the farthest node and its distance from the start node
    pair<int, int> findFarthestNode(int start, const vector<vector<int>>& adj) {
        vector<int> dist(adj.size(), -1);
        queue<int> q;
        q.push(start);
        dist[start] = 0;

        int farthest_node = start;
        int max_dist = 0;

        while (!q.empty()) {
            int node = q.front();
            q.pop();
            
            for (int neighbor : adj[node]) {
                if (dist[neighbor] == -1) {
                    dist[neighbor] = dist[node] + 1;
                    q.push(neighbor);
                    if (dist[neighbor] > max_dist) {
                        max_dist = dist[neighbor];
                        farthest_node = neighbor;
                    }
                }
            }
        }

        return {farthest_node, max_dist};
    }

    // BFS to get distance from start node to all other nodes
    vector<int> bfs(int start, const vector<vector<int>>& adj) {
        vector<int> dist(adj.size(), -1);
        queue<int> q;
        q.push(start);
        dist[start] = 0;

        while (!q.empty()) {
            int node = q.front();
            q.pop();
            
            for (int neighbor : adj[node]) {
                if (dist[neighbor] == -1) {
                    dist[neighbor] = dist[node] + 1;
                    q.push(neighbor);
                }
            }
        }

        return dist;
    }
};

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