<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 Node {
public:
    int val;
    unordered_set<Node*> neighbor;
    Node(int v) : val(v) {};
};

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<vector<int>>& edges) {
        // onion peeling, the left 1 or 2 are the result
        // peeling: start from leaves, once remove a leaf, try remove its parent if becomes a leaf; so we should mark parents when getting leaves;
        if (n == 1) return {0};
        if (n == 2) return {0, 1};

        vector<Node*> tree(n, nullptr);

        for (vector<int> edge : edges) {
            int p = edge[0], c = edge[1];
            if (tree[p] == nullptr) tree[p] = new Node(p);
            if (tree[c] == nullptr) tree[c] = new Node(c);

            tree[p]->neighbor.insert(tree[c]);
            tree[c]->neighbor.insert(tree[p]);
        }

        // start from leaves
        queue<pair<Node*, int>> levels;
        for (Node* node : tree) {
            if (node->neighbor.size() == 1) {
                levels.push({node, 0});
            }
        }

        int remaining = n, currentLevel = 0;;

        while (!levels.empty()) {
            pair<Node*, int> p = levels.front();
            Node* node = p.first;
            levels.pop();

            Node* parent = *(node->neighbor.begin()); // only 1 element in the set
            parent->neighbor.erase(node);
            if (parent->neighbor.size() == 1) {
                levels.push({parent, p.second + 1});
            }

            tree[node->val] = nullptr;
            remaining--;
            currentLevel = p.second;

            if (remaining <= 2) break;
        }

        if (levels.front().second == currentLevel) {
            tree[levels.front().first->val] = nullptr;
        }

        vector<int> result;
        for (int i = 0; i < n; i++) {
            if (tree[i] != nullptr) result.push_back(i);
        }

        return result;
    }
};
91990