310. Minimum-Height-Trees

difficulty: Medium

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.

Given a tree of n nodes labelled from 0 to n - 1, and an array of n - 1 edges where edges[i] = [ai, bi] indicates that there is an undirected edge between the two nodes ai and bi in the tree, you can choose any node of the tree as the root. When you select a node x as the root, the result tree has height h. Among all possible rooted trees, those with minimum height (i.e. min(h)) are called minimum height trees (MHTs).

Return a list of all MHTs' root labels. You can return the answer in any order.

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

Example 1:

Input: n = 4, edges = [[1,0],[1,2],[1,3]]
Output: [1]
Explanation: As shown, the height of the tree is 1 when the root is the node with label 1 which is the only MHT.

Example 2:

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

Example 3:

Input: n = 1, edges = []
Output: [0]

Example 4:

Input: n = 2, edges = [[0,1]]
Output: [0,1]

Constraints:

• 1 <= n <= 2 * 104

• edges.length == n - 1

• 0 <= ai, bi < n

• ai != bi

• All the pairs (ai, bi) are distinct.

• The given input is guaranteed to be a tree and there will be no repeated edges.

Method One

class Solution {
    /*
    这个讨论的最高评论直接杀死比赛，解释了一切。
    https://leetcode.com/problems/minimum-height-trees/discuss/76055/Share-some-thoughts
    我们用 indegrees 找到所有的 leaf, 然后同时开始BFS。
    我们每次记录上一层BFS 的 leaves, 这样我们最后一层 BFS 的 leaves 就是所有的正确答案了。
    */
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        if( n == 0){
            return new ArrayList<Integer>();
        }
        if( n == 1 ){
            ArrayList<Integer> temp = new ArrayList<Integer>();
            temp.add(0);
            return temp;
        }
        int[] degrees = new int[n];
        ArrayList<Integer>[] graph = new ArrayList[n];
        for(int i = 0; i < n; i++ ){
            graph[i] = new ArrayList<>();
        }
        
        for(int[] edge : edges){
            int a = edge[0];
            int b = edge[1];
            graph[a].add(b);
            graph[b].add(a);
            degrees[a] += 1;
            degrees[b] += 1;
        }
        LinkedList<Integer> bfs = new LinkedList<>();
        boolean[] marked = new boolean[n];
        
        for(int i = 0; i < n; i++ ){
            if( degrees[i] == 1 ){
                bfs.offer(i);
            }
        }
        
        List<Integer> prevLeaves = new ArrayList<>();
        while( !bfs.isEmpty() ){
            int size = bfs.size();
            prevLeaves = new ArrayList<>(bfs);
            while( size > 0 ){
                size -= 1;
                int cur = bfs.poll();
                marked[cur] = true;
                for( int next  : graph[cur] ){
                    if( marked[next] ){
                        continue;
                    }
                    degrees[next] -= 1;
                    if( degrees[next] == 1){
                        bfs.offer(next);
                    }
                }
            }            
        }
        return prevLeaves;
    }
}
​

上面的答案在2023年被证实有缺陷。因为没有考虑 indegrees == 1 的节点。

class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer>[] adjList = new ArrayList[n];
        int[] indegrees = new int[n];
        for (int i = 0; i < n; i++) {
            adjList[i] = new ArrayList<>();
        }
        for (int[] edge : edges) {
            adjList[edge[0]].add(edge[1]);
            adjList[edge[1]].add(edge[0]);
            indegrees[edge[0]]++;
            indegrees[edge[1]]++;
        }
        List<Integer> ans = new ArrayList<>();
        ArrayDeque<Integer> queue = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            if (indegrees[i] == 1) {
                queue.offer(i);
            }
            if (indegrees[i] == 0) {
                ans.add(i);
            }
        }
        ArrayList<Integer> prevLevel = new ArrayList<>();
        while (!queue.isEmpty()) {
            int size = queue.size();
            prevLevel = new ArrayList<>(queue);
            while (size > 0) {
                int cur = queue.poll();
                for (int next : adjList[cur]) {
                    indegrees[next]--;
                    if (indegrees[next] == 1) {
                        queue.offer(next);
                    }
                }
                size--;
            }
        }
        ans.addAll(prevLevel);
        return ans;
    }
}