<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) {
            vector<int> tmp({0});
            return tmp;
        }
        if (n == 2) {
            return edges[0];
        }
        unordered_map<int, set<int>> treeMap;
        for (auto& vec : edges) {
            treeMap[vec[0]].emplace(vec[1]);
            treeMap[vec[1]].emplace(vec[0]);
        }

        //find all tree nodes
        int bigSize = 1;
        

        do {
            //cout << "loop" << endl;
            bigSize = 1;
            vector<int> toErase;
            vector<vector<int>> nErase;
            for (auto& pair : treeMap) {
                // finds the current children nodes
                if (pair.second.size() == 1) {
                    //cout << "child node: " << pair.first << endl;
                    int neighborVal = *pair.second.begin();
                    int nodeVal = pair.first;
                    // remove the node val itself
                    pair.second.erase(pair.second.begin());
                    toErase.push_back(pair.first);
                    // remove the node from consideration;
                    // remove the node from the neighbor val
                    nErase.push_back({ neighborVal, nodeVal });
                    
                    // update size
                    //bigSize = max(int(treeMap[neighborVal].size()), bigSize);
                }
            }
            //cout << "erasing neighbors" << endl;
            // erase in neighbors
            cout << nErase.size() << endl;
            for (auto& val : nErase) {
                int neighborVal = val[0];
                int nodeVal = val[1];
                if (treeMap[neighborVal].find(nodeVal) != treeMap[neighborVal].end()) 
                    treeMap[neighborVal].erase(treeMap[neighborVal].find(nodeVal));
            }
            //cout << "erasing vals" << endl;
            // erase vals
            for (int key : toErase) {
                //cout << "erase: " << key << endl;
                treeMap.erase(key);
            }
            //cout << "finding max" << endl;
            for (auto& pair : treeMap) {
                bigSize = max(bigSize, int(pair.second.size()));
            }
            //cout << bigSize << endl;
        } while (bigSize > 1);

        // find current nums
        vector<int> result;
        for (auto& pair : treeMap) {
            result.push_back(pair.first);
        }
        return result;

    }
    
};
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