Minimum Height Trees

Description

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 contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).

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

Example 1:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

    0
    |
    1
   / \
  2   3

return [1]

Example 2:

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

 0  1  2
  \ | /
    3
    |
    4
    |
    5

return [3, 4]

Hint

Train of Thought

对一个tree in graph来说, 它的高度就是它的longest path, 想要让最长path最短, 就只有中点, 否则离左边近, 离右边就远

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.

Code

public List<Integer> findMinHeightTrees(int n, int[][] edges) {
    if (n == 1) return Collections.singletonList(0);

    List<Set<Integer>> adj = new ArrayList<>(n);
    for (int i = 0; i < n; ++i) adj.add(new HashSet<>());
    for (int[] edge : edges) {
        adj.get(edge[0]).add(edge[1]);
        adj.get(edge[1]).add(edge[0]);
    }

    List<Integer> leaves = new ArrayList<>();
    for (int i = 0; i < n; ++i)
        if (adj.get(i).size() == 1) leaves.add(i);

    while (n > 2) {
        n -= leaves.size();
        List<Integer> newLeaves = new ArrayList<>();
        for (int leave : leaves) {
            int fatherOfLeave = adj.get(leave).iterator().next();
            adj.get(fatherOfLeave).remove(leave);
            if (adj.get(fatherOfLeave).size() == 1) newLeaves.add(fatherOfLeave);
        }
        leaves = newLeaves;
    }
    return leaves;
}

Complexity

time: O(n) space: O(n)