leetcode：Minimum Height Trees

最新推荐文章于 2025-12-26 21:45:18 发布

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#bfs #leetcode #遍历

本文介绍了一种使用广度优先搜索(BFS)策略找到给定无向图中所有最小高度树的方法。通过逐步移除度为1的节点直至剩下1或2个节点作为根节点，实现对最小高度树的有效寻找。

先上题目：

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]]

return [1]

Example 2:

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

return [3, 4]

分析：

首先排除特殊情况，即只有1个结点或无结点。

然后考虑一般情况，这里结合上述的2个example，我觉得可以借助BFS，只要我们多次遍历，不停地删除页结点（度为1的结点），到最后剩余1或2结点时，那边是我们所求的答案。代码如下：

class Solution {
public:
vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
    // 初始化边
    vector<unordered_set<int>> adj(n);
    for (pair<int, int> p : edges) {
      adj[p.first].insert(p.second);
      adj[p.second].insert(p.first);
    }
    // 仅一个结点的情况
    vector<int> current;
    if (n == 1) {
      current.push_back(0);
      return current;
    }
    // 寻找第一批叶子结点
    for (int i = 0; i < adj.size(); ++i) {
      if (adj[i].size() == 1) {
        current.push_back(i);
      }
    }
    // BFS
    while (true) {
      vector<int> next;
      for (int node : current) {
        for (int neighbor : adj[node]) {
          adj[neighbor].erase(node);
          if (adj[neighbor].size() == 1) next.push_back(neighbor);
        }
      }
      if (next.empty()) return current;
      current = next;
    }
}
};