310. Minimum Height Trees

題目：

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.

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

題解：

求給定圖中，能形成樹的最矮的樹。第一直覺就是BFS，跟Topological Sorting的Kahn方法很類似，利用無向圖每個點的degree來計算。但是卻后繼無力，於是還是參考了Discuss中Dietpepsi和Yavinci大神的代碼。

方法有兩種，一種是先計算每個點的degree，然后將degree為1的點放入list或者queue中進行計算，把這些點從neighbours中去除，然后計算接下來degree = 1的點。最后剩下1 - 2個點就是新的root

另外一種是用了類似給許多點，求一個點到其他點距離最短的原理。找到最長的一點leaf to leaf path，然后找到這條path的一個或者兩個中點median就可以了。

下面是用第一種方法做的。

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;
    }
}

 

題外話：

今天下午得知群里好幾個都是caltech的大神...拜一拜，拜一拜

 

二刷:

兩種方法，一種是無向圖中求longest path，假如longest path長度是奇數，則結果為最中間的一個節點，否則為最中間的兩個節點。思路好想到，但是並不好寫。求Longest Path是一個NP-Hard問題。但對於DAG來說可以用dp來求出結果。這里un-directed graph我們也可以試一試。

另一種方法是用BFS類似Topological Sorting中的Kahn方法。先計算每個節點的degree，然后把低degree的節點leaf放入queue中進行處理，一層一層把低degree節點逐漸剝離，最后剩下的1 - 2個節點就是解。

Java:

超時的longest path找中點

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;
    }
}

 

參考樂神和Yavinci的remove leaf:

跟一刷一樣。先計算degree為1的節點，這些節點只和一個節點相連，所以這些是leaf節點。逐個去除掉leaf節點以后我們可以嘗試計算上一層leaf，繼續and繼續，直到最后我們剩下一個節點或者兩個節點，就是我們要求的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/