[LeetCode] 310. Minimum Height Trees_Medium tag: BFS

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 :

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 exactlyone 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.

这个题目思路来自于Solution, 类似于剥洋葱, 从leaves开始一层一层往里面剥, 依次更新leaves, 到最后只剩一个或2个nodes的时候(因为根据tree的定义, 从数学上可以证明最多只有两个root的height一样), 那么剩下的点就是答案.

1. Constraints

1) 题目意思已经很清楚了, 只是需要判断edge case: n = 1, n= 2 时

2. Ideas

类似于BFS的解法      T: O(n)    S: O(n)

3. Code

class Solution:
    def miniHeightTree(self, n, edges):
        if n <3: return [i for i in range(n)]
        graph = collections.defaultdict(set)
        for c1, c2 in edges:
            graph[c1].add(c2)
            graph[c2].add(c1)
        leaves = [i for i in range(n) if len(graph[i]) == 1]
        while n > 2:
            n -= len(leaves)
            newleaves = []
            for i in leaves: #take off the outside leaves
                neig = graph[i].pop() # neighbour of the leave
                graph[neig].remove(i) #remove the edge from leave to neighbor
                if len(graph[neig]) == 1:
                    newleaves.append(neig)
            leaves = newleaves
        return leaves