# 310 Minimum Height Trees

## 310. [Minimum Height Trees](https://leetcode.com/problems/minimum-height-trees/description/)

## 1. Question

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](https://en.wikipedia.org/wiki/Tree_%28graph_theory%29): “a tree is an undirected graph in which any two vertices are connected byexactlyone 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.

## 2. Implementation

**(1) BFS**

```java
class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer> res = new ArrayList<>();

        if (n == 1) {
            res.add(0);
            return res;
        }

        List<Set<Integer>> adjList = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            adjList.add(new HashSet<>());
        }

        for (int[] edge : edges) {
            adjList.get(edge[0]).add(edge[1]);
            adjList.get(edge[1]).add(edge[0]);
        }

        List<Integer> leaves = new ArrayList<>();
        List<Integer> newLeaves = null;

        for (int i = 0; i < n; i++) {
            if (adjList.get(i).size() == 1) {
                leaves.add(i);
            }
        }

        while (n > 2) {
            n -= leaves.size();
            newLeaves = new ArrayList<>();

            for (int node : leaves) {
                for (int parent : adjList.get(node)) {
                    adjList.get(parent).remove(node);
                    if (adjList.get(parent).size() == 1) {
                        newLeaves.add(parent);
                    }
                }
            }
            leaves = newLeaves;
        }
        return leaves;
    }
}
```

## 3. Time & Space Complexity

BFS: 时间复杂度O(n)， 空间复杂度O(n)


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