Minimum Height Trees
A tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.
Given a tree of n nodes labelled from 0 to n - 1, and an array of n - 1 edges where edges[i] = [ai, bi] indicates that there is an undirected edge between the two nodes ai and bi in the tree, you can choose any node of the tree as the root. When you select a node x as the root, the result tree has height h. Among all possible rooted trees, those with minimum height (i.e. min(h)) are called minimum height trees (MHTs).
Return a list of all MHTs' root labels. You can return the answer in any order.
The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.
Input: n = 4, edges = [[1,0],[1,2],[1,3]]
Output: [1]
Explanation: As shown, the height of the tree is 1 when the root is the node with label 1 which is the only MHT.As the hints suggest, this problem is related to the graph data structure. Moreover, it is closely related to the problems of Course Schedule and Course Schedule II. This relationship is not evident, yet it is the key to solve the problem, as one will see later.
First of all, as a straight-forward way to solve the problem, we can simply follow the requirements of the problem, as follows:
Starting from each node in the graph, we treat it as a root to build a tree. Furthermore, we would like to know the distance between this root node and the rest of the nodes. The maximum of the distance would be the height of this tree.
Then according to the definition of Minimum Height Tree (MHT), we simply filter out the roots that have the minimal height among all the trees.
The first step we describe above is actually the problem of Maximum Depth of N-ary Tree, which is to find the maximum distance from the root to the leaves nodes. For this, we can either apply the Depth-First Search (DFS) or Breadth-First Search (BFS) algorithms.
Without a rigid proof, we can see that the above straight-forward solution is correct, and it would work for most of the test cases.
However, this solution is not efficient, whose time complexity would be \mathcal{O}(N^2)O(N2) where NN is the number of nodes in the tree. As one can imagine, it will result in Time Limit Exceeded exception in the online judge.
As a spoiler alert, in this article, we will present a topological sorting alike algorithm with time complexity of O(n), which is also the algorithm to solve the well-known course schedule problems.
Approach 1
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