310. Minimum Height Trees

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]]

return [1]

Example 2:

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

return [3, 4]

Hint:

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

二刷:

两种方法，一种是无向图中求longest path，假如longest path长度是奇数，则结果为最中间的一个节点，否则为最中间的两个节点。思路好想到，但是并不好写。求Longest Path是一个NP-Hard问题。但对于DAG来说可以用dp来求出结果。这里un-directed graph我们也可以试一试。

另一种方法是用BFS类似Topological Sorting中的Kahn方法。先计算每个节点的degree，然后把低degree的节点leaf放入queue中进行处理，一层一层把低degree节点逐渐剥离，最后剩下的1 - 2个节点就是解。

Java:

超时的longest path找中点

Time Complexity - O(n²)， Space Complexity - O(n) <- TLE

public class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer> res = new ArrayList<>();
        if (edges == null || edges.length == 0 || edges.length != n - 1) return res;
        List<LinkedList<Integer>> paths = new ArrayList<>();    // no cycle, no duplicate
        int len = 0, maxIndex = 0;
        for (int[] edge : edges) {
            for (int i = 0; i < paths.size(); i++) {
                LinkedList<Integer> path = paths.get(i);
                if (path.peekFirst() == edge[0]) path.addFirst(edge[1]);
                else if (path.peekFirst() == edge[1]) path.addFirst(edge[0]);
                else if (path.peekLast() == edge[0]) path.addLast(edge[1]);
                else if (path.peekLast() == edge[1]) path.addLast(edge[0]);
                
                if (paths.get(i).size() > len) {
                    len = paths.get(i).size();
                    maxIndex = i;
                }
            }
            paths.add(new LinkedList<>(Arrays.asList(new Integer[] {edge[0], edge[1]})));
        }
        
        LinkedList<Integer> longestPath = paths.get(maxIndex);
        if (longestPath.size() % 2 == 0) {
            res.add(longestPath.get(longestPath.size() / 2 - 1));
            res.add(longestPath.get(longestPath.size() / 2));
        } else {
            res.add(longestPath.get(longestPath.size() / 2));
        }
        return res;
    }
}

参考乐神和Yavinci的remove leaf:

跟一刷一样。先计算degree为1的节点，这些节点只和一个节点相连，所以这些是leaf节点。逐个去除掉leaf节点以后我们可以尝试计算上一层leaf，继续and继续，直到最后我们剩下一个节点或者两个节点，就是我们要求的root nodes。

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);
        List<Set<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) graph.add(new HashSet<>());
        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 j : graph.get(leaf)) {
                    graph.get(j).remove(leaf);
                    if (graph.get(j).size() == 1) newLeaves.add(j);
                }
            }
            leaves = newLeaves;
        }
        return leaves;
    }
}

Reference:

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

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

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

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

https://discuss.codechef.com/questions/51180/finding-longest-path-in-an-undirected-and-unweighted-graph

http://cs.nyu.edu/courses/spring14/CSCI-UA.0480-004/Lecture14.pdf

http://www.geeksforgeeks.org/find-longest-path-directed-acyclic-graph/

posted @ 2015-12-20 06:41 YRB 阅读(2351) 评论(2) 编辑收藏举报

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310. Minimum Height Trees

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