<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:
    def findMinHeightTrees(self, n: int, edges: List[List[int]]) -> List[int]:
        # Last nodes in topological sort will be most connected and have MHTs
        # At most, 2 distinct nodes with MHTs (think of nodes in straight line - odd/even)
        # BFS to compare MHTs starting at last 2 nodes of topological sort
        # O(V) time, O(V) space
        
        adj_list = {node: [] for node in range(n)}
        for src, dest in edges:
            adj_list[src].append(dest)
            adj_list[dest].append(src)
                
        def topological_sort():
            in_degrees = {node: 0 for node in range(n)}
            for src_node in adj_list:
                for dst_node in adj_list[src_node]:
                    in_degrees[dst_node] += 1

            queue = collections.deque([node for node in in_degrees if in_degrees[node] == 1])
            topological_order = []
            while queue:
                node = queue.popleft()
                topological_order.append(node)
                for neighbor in adj_list[node]:
                    in_degrees[neighbor] -= 1
                    if in_degrees[neighbor] == 1:
                        queue.append(neighbor)
            if len(topological_order) >= 2:
                return topological_order[-2:]
            else:
                return topological_order
        
        def bfs(root):
            queue = collections.deque([root])
            visited = set([root])
            distance = 0
            while queue:
                for _ in range(len(queue)):
                    node = queue.popleft()
                    for neighbor in adj_list[node]:
                        if neighbor not in visited:
                            queue.append(neighbor)
                            visited.add(neighbor)
                distance += 1
            return distance
        
        if n == 1:
            return [0]
        topological_order = topological_sort()
        if bfs(topological_order[0]) == bfs(topological_order[1]):
            return topological_order
        else:
            return topological_order[1:]
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