Leetcode: Minimum Height Trees

Minimum Height Trees



Similar Problems:


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]

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


// Blog link: https://code.dennyzhang.com/minimum-height-trees
// Basic Ideas:
// There are at most 2 MHTs
// We keep removing leaf nodes. The remaining one or two nodes are targets
// Complexity: Time O(n), Space O(n)
func findMinHeightTrees(n int, edges [][]int) []int {
    if n == 1 { return []int{0} }
    relations := make([]map[int]bool, n)
    for i:= 0; i<n; i++ { relations[i] = map[int]bool{}}
    for _, edge := range edges {
        p, q := edge[0], edge[1]
        relations[p][q], relations[q][p] = true, true
    }
    queue := []int{}
    // start with leaf nodes
    for i:=0; i<n; i++ {
        if len(relations[i]) == 1 { queue = append(queue, i) }
    }

    for n>2 {
        // explore current level
        n -= len(queue)
        items := []int{}
        for _, node := range queue {
            for node2 := range relations[node] {
                delete (relations[node2], node)
                // inner layer
                if len(relations[node2]) == 1 { items = append(items, node2) }
            }
        }
        queue = items
    }
    return queue
}
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