leetcode310. Minimum Height Trees

xiaoxiaozi 發(fā)布于2019-08-16 13:46 / 596人閱讀

摘要：現(xiàn)在要求在這樣一棵生成樹中，找到生成樹的高度最低的所有根節(jié)點。然后利用鄰接表的相關(guān)屬性來判斷當(dāng)前節(jié)點是否是葉節(jié)點。度數(shù)為一的頂點就是葉節(jié)點。這里使用異或的原因是對同一個值進行兩次異或即可以回到最初值。

題目

For an 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 :

Input: n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / 
      2   3 

Output: [1]
Example 2 :

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

     0  1  2
       | /
        3
        |
        4
        |
        5 

Output: [3, 4]
Note:

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

在無向圖的生成樹中，我們可以指定任何一個節(jié)點為這棵樹的根節(jié)點。現(xiàn)在要求在這樣一棵生成樹中，找到生成樹的高度最低的所有根節(jié)點。

其實，決定一棵樹的高度往往是樹中的最長路徑，只有我們選擇最長路徑的中間點才能夠盡可能減少樹的高度。那么我們?nèi)绾握业剿械闹虚g節(jié)點呢？

當(dāng)出發(fā)點只有兩個時，我們知道中間節(jié)點就是從出發(fā)點同時出發(fā)，當(dāng)二者相遇或者二者只間隔一個位置是所在的點就是兩個出發(fā)點的重點。那么多個出發(fā)點也是同樣的道理，每個人從各個出發(fā)點出發(fā)，最終相遇的點就是我們所要找的中間點。

這題的思路有些類似于拓?fù)渑判颍看挝覀兌紩コ腥攵葹?的點，因為這些點肯定是葉節(jié)點。然后不停的往中間走，直到剩下最后一個葉節(jié)點或是最后兩個葉節(jié)點。

這個圖中刪除所有入度為0的點就只剩下1，因此我們知道1一定就是我們所要求的根節(jié)點

思路一：圖論

這一種解法著重強調(diào)了利用圖論中的數(shù)據(jù)結(jié)構(gòu)來解決問題。這里我們采用圖論中的鄰接表來存儲圖中的點和邊。然后利用鄰接表的相關(guān)屬性來判斷當(dāng)前節(jié)點是否是葉節(jié)點。

    public List findMinHeightTrees(int n, int[][] edges) {
        if(n==1) return Collections.singletonList(0);
        //初始化鄰接表
        List> adj = new ArrayList>();
        for(int i = 0 ; i());
        }
        for(int[] edge : edges) {
            adj.get(edge[0]).add(edge[1]);
            adj.get(edge[1]).add(edge[0]);
        }
        
        List leaves = new ArrayList();
        for(int i = 0 ; i 2) {
            n -= leaves.size();
            List newLeaves = new ArrayList<>();
            for (int i : leaves) {
                int j = adj.get(i).iterator().next();
                adj.get(j).remove(i);
                if (adj.get(j).size() == 1) newLeaves.add(j);
            }
            leaves = newLeaves;
        }
        return leaves;
    }

思路二：簡化數(shù)據(jù)結(jié)構(gòu)

這里使用degree數(shù)組存儲每個頂點的度數(shù)，即連接的變數(shù)。度數(shù)為一的頂點就是葉節(jié)點。再用connected存儲每個頂點所連接的所有邊的異或值。這里使用異或的原因是對同一個值進行兩次異或即可以回到最初值。

    public List findMinHeightTrees2(int n, int[][] edges) {
        if(n==1) return Collections.singletonList(0);
        int[] connected = new int[n];
        int[] degree = new int[n];
        
        for(int[] edge : edges) {
            int v1 = edge[0];
            int v2 = edge[1];
            connected[v1] ^= v2;
            connected[v2] ^= v1;
            
            degree[v1]++;
            degree[v2]++;
        }
        
        LinkedList queue = new LinkedList();
        for(int i = 0 ; i 2 && !queue.isEmpty()) {
            int size = queue.size();
            for(int i = 0 ; i result = new ArrayList();
        result.addAll(queue);
        return result;
    }

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