<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:
    int maxDepth;
    int farthestNode;
    void dfs(int s, vector<bool> &visited, vector< vector<int> > &adj, int currDepth, vector<int> &path) {
        visited[s] = true;
        if(maxDepth < currDepth) {
            maxDepth = currDepth;
            farthestNode = s;
        }
        for(auto i : adj[s]) {
            if(!visited[i]) {
                path[i] = s;
                dfs(i, visited, adj, currDepth + 1, path);
            }
        }
    }
    
    vector<int> findMinHeightTrees(int n, vector<vector<int>>& edges) {
        // find any leaf node and store the nodes based on the height
        // while finding the diameter of the tree
        // output the nodes with middle height
        vector< vector<int> > adj(n, vector<int>(0));
        for(int i = 0; i < edges.size(); i++) {
            adj[edges[i][0]].push_back(edges[i][1]);
            adj[edges[i][1]].push_back(edges[i][0]);
        }
        
        // resetting values
        
        farthestNode = 0;
        maxDepth = 0;
        vector<int> path(n, -1);
        vector<bool> visited(n, false);
        
        // first DFS to find any leaf node
        dfs(0, visited, adj, 0, path);
        
        // resetting the values for next DFS which finds the diameter and its path
        maxDepth = -1;
        fill(visited.begin(), visited.end(), false);
        fill(path.begin(), path.end(), -1);
        dfs(farthestNode, visited, adj, 0, path);
        
        // finding the middle node by traversing the path
        int u = farthestNode;
        int currDepth = 0;
        vector<int> result;
        while(u != -1) {
            if(maxDepth % 2) {
                if(currDepth == (maxDepth / 2) || currDepth == ((maxDepth + 1) / 2)) {
                    result.push_back(u);
                }
            } else {
                if(currDepth == (maxDepth / 2)) {
                    result.push_back(u);
                }
            }
            currDepth++;
            u = path[u];
        }
        return result;
    }
};
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