找到树的父节点以创建尽可能短的树高
我有一个无向图，表示为欧几里德权重的邻接矩阵。我用它来表示更大完整图的最小生成树。
我想要找到的是图中的单个节点，当用作根节点时，会创建尽可能短的树高度。我想出的是使用每个节点作为根执行深度优先遍历，并跟踪所看到的最短高度。有没有更快的方法来完成这个任务？
I have an undirected graph represented as an adjacency matrix of Euclidean weights. I'm using this to represent the minimum spanning tree for a larger complete graph.
What I want to find is the single node within the graph that, when used as the root node, creates the shortest possible tree height. What I've come up with is performing a depth-first traversal using every node as the root, and keeping track of the shortest height seen. Is there a faster way to accomplish this?
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这是一道经典的算法题。您要寻找的东西称为树的中心，可以使用简单的迭代算法找到它。 这个问题有一个很好的答案，解释了如何做到这一点。
希望这有帮助！
This is a classic algorithms question. What you're looking for is called the center of the tree, and it can be found using a simple iterative algorithm. This question has a great answer that explains how to do it.
Hope this helps!