310. Minimum Height Trees

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题目：

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:

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

Note:

(1) According to the definition of tree on Wikipedia: “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.”

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

链接： http://leetcode.com/problems/minimum-height-trees/

题解：

求给定图中，能形成树的最矮的树。第一直觉就是BFS，跟Topological Sorting的Kahn方法很类似，利用无向图每个点的degree来计算。但是有却后继无力，于是还是参考了Discuss中Dietpepsi和Yavinci大神的代码。

方法有两种，一种是先计算每个点的degree，然后将degree为1的点放入list或者queue中进行计算，把这些点从neighbours中去除，然后计算接下来degree = 1的点。最后剩下1 - 2个点就是新的root

另外一种是用了类似给许多点，求一个点到其他点距离最短的原理。找到最长的一点leaf to leaf path，然后找到这条path的一个或者两个中点median就可以了。

下面是用第一种方法做的。

Time Complexity - O(n)， Space Complexity - O(n)

public class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer> leaves = new ArrayList<>();
        if(n <= 1) {
            return Collections.singletonList(0);
        }
        Map<Integer, Set<Integer>> graph = new HashMap<>();     // list of edges to  Ajacency Lists
        
        for(int i = 0; i < n; i++) {
            graph.put(i, new HashSet<Integer>());
        }
        for(int[] edge : edges) {
            graph.get(edge[0]).add(edge[1]);
            graph.get(edge[1]).add(edge[0]);
        }
        
        for(int i = 0; i < n; i++) {
            if(graph.get(i).size() == 1) {
                leaves.add(i);
            }
        }
        
        while(n > 2) {
            n -= leaves.size();
            List<Integer> newLeaves = new ArrayList<>();
            for(int leaf : leaves) {
                for(int newLeaf : graph.get(leaf)) {
                    graph.get(leaf).remove(newLeaf);
                    graph.get(newLeaf).remove(leaf);
                    if(graph.get(newLeaf).size() == 1) {
                        newLeaves.add(newLeaf);
                    }
                }
            }
            leaves = newLeaves;
        }
        
        return leaves;
    }
}

题外话：

今天下午得知群里好几个都是caltech的大神...拜一拜，拜一拜

Reference:

https://leetcode.com/discuss/71763/share-some-thoughts

https://leetcode.com/discuss/71738/easiest-75-ms-java-solution

https://leetcode.com/discuss/71656/c-solution-o-n-time-o-n-space

https://leetcode.com/discuss/72739/two-o-n-solutions

https://leetcode.com/discuss/71804/java-layer-by-layer-bfs

https://leetcode.com/discuss/71721/iterative-remove-leaves-python-solution

https://leetcode.com/discuss/71802/solution-share-midpoint-of-longest-path

https://leetcode.com/discuss/73926/share-java-using-degree-with-explanation-which-beats-more-than

https://leetcode.com/discuss/71676/share-my-accepted-solution-java-o-n-time-o-n-space

310. Minimum Height Trees

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原文地址：http://www.cnblogs.com/yrbbest/p/5060225.html