310. Minimum Height Trees (Medium)

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format

The graph containsnnodes which are labeled from0ton - 1. You will be given the numbernand a list of undirectededges(each edge is a pair of labels).

You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges.

Example 1:

Givenn = 4,edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3

return[1]

Example 2:

Givenn = 6,edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
      \ | /
        3
        |
        4
        |
        5

return[3, 4]

Hint:

1. How many MHTs can a graph have at most?

Note:

(1) According to thedefinition of tree on Wikipedia): “a tree is an undirected graph in which any two vertices are connected byexactlyone path. In other words, any connected graph without simple cycles is a tree.”

(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

Solution: BFS topological sort

Time Complexity: $$O(n)$$ 96ms

Idea:

First let's review some statement for tree in graph theory:

(1) A tree is an undirected graph in which any two vertices are
connected by exactly one path.

(2) Any connected graph who hasnnodes withn-1edges is a tree.

(3) The degree of a vertex of a graph is the number of
edges incident to the vertex.

(4) A leaf is a vertex of degree 1. An internal vertex is a vertex of
degree at least 2.

(5) A path graph is a tree with two or more vertices that is not
branched at all.

(6) A tree is called a rooted tree if one vertex has been designated
the root.

(7) The height of a rooted tree is the number of edges on the longest
downward path between root and a leaf.

OK. Let's stop here and look at our problem.

Our problem want us to find the minimum height trees and return their root labels. First we can think about a simple case -- a path graph.

For a path graph of n nodes, find the minimum height trees is trivial. Just designate the middle point(s) as roots.

Despite its triviality, let design a algorithm to find them.

Suppose we don't know n, nor do we have random access of the nodes. We have to traversal. It is very easy to get the idea of two pointers. One from each end and move at the same speed. When they meet or they are one step away, (depends on the parity of n), we have the roots we want.

This gives us a lot of useful ideas to crack our real problem.

For a tree we can do some thing similar. We start from every end, by end we mean vertex of degree 1 (aka leaves). We let the pointers move the same speed. When two pointers meet, we keep only one of them, until the last two pointers meet or one step away we then find the roots.

It is easy to see that the last two pointers are from the two ends of the longest path in the graph.

The actual implementation is similar to the BFS topological sort. Remove the leaves, update the degrees of inner vertexes. Then remove the new leaves. Doing so level by level until there are 2 or 1 nodes left. What's left is our answer!

The time complexity and space complexity are both O(n).

Note that for a tree we always have V = n, E = n-1.

vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
    if (n == 1) return {0};
    vector<int> res;
    vector<set<int>> adj(n);
    queue<int> q;
    for (auto edge: edges) {
        adj[edge.first].insert(edge.second);
        adj[edge.second].insert(edge.first);
    }

    for (int i = 0; i < n; ++i) {
        if (adj[i].size() == 1) q.push(i);
    }

    while (n > 2) {
        int size = q.size();
        n -= size;
        for (int i = 0; i < size; ++i) {
            int t = q.front(); q.pop();
            for (auto a:adj[t]) {
                adj[a].erase(t);
                if (adj[a].size() == 1) q.push(a);
            }
        }
    }

    while (!q.empty()) {
        res.push_back(q.front()); q.pop();
    }
    return res;
}

Solution 2:

Time Complexity: $$O(n)$$

The basic idea is to calculate the longest path for each node with only $$O(3n)$$ scan.

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