<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 i:edges)
        {
            int u = i[0] , v = i[1];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        vector<int> vis(n,0);
        function<int(int )>bfs = [&](int node)
        {
            queue<int> q;
            vis[node]=1;
            q.push(node);
            int ans=node;
            while(!q.empty())
            {
                int curr = q.front();
                ans = curr;
                q.pop();
                 for(auto i:adj[curr])
                 {
                    if(!vis[i])
                    {
                        vis[i]=1;
                        q.push(i);
                    }
                 }
            }
            return ans;
        };
        vector<int> sz(n,0),par(n,0);
        // vis.resize(n, 0);  // Resize to new_size and set all elements to 0

        function<void(int,int)>dfs = [&](int node ,int height)
        {
            cout<<node<<endl;
            // cout<<vis[node]<<endl;
            vis[node] = 1;
            sz[node] = height;
            for(auto i:adj[node])
            {
                if(!vis[i])
                {
                    par[i] = node;
                    dfs(i,height+1);
                }
            }   
        };
        int node = bfs(0);
        // cout<<node<<endl;
        par[node] = node;
                for (auto& element : vis) {
            element = 0;
}
        dfs(node,0);
        int maxm = max_element(sz.begin(), sz.end()) - sz.begin();

        int hmax = sz[maxm] / 2;
        int hmax2 = ceil((double)sz[maxm] / 2);
        int curr = maxm;
        int x = 0;
        int ans1 ,ans2 = -1;

        while(par[curr]!=curr)
        {
            ans1 = curr;
            if(x == hmax)
            {
                if(hmax != hmax2)
                {
                    ans2 = par[curr];
                }
                break;
                
            }
            curr = par[curr];
            x++;
            
        }
        // cout<<node<<endl;
        if(ans2 == -1)
        return {ans1};
        return {ans1,ans2};

    }
};
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