0310. 最小高度树【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - 拓扑排序剥叶

1. 📝 题目描述

• leetcode

树是一个无向图，其中任何两个顶点只通过一条路径连接。 换句话说，任何一个没有简单环路的连通图都是一棵树。

给你一棵包含 n 个节点的树，标记为 0 到 n - 1。给定数字 n 和一个有 n - 1 条无向边的 edges 列表（每一个边都是一对标签），其中 edges[i] = [ai, bi] 表示树中节点 ai 和 bi 之间存在一条无向边。

可选择树中任何一个节点作为根。当选择节点 x 作为根节点时，设结果树的高度为 h。在所有可能的树中，具有最小高度的树（即，min(h)）被称为 最小高度树。

请你找到所有的 最小高度树 并按 任意顺序 返回它们的根节点标签列表。

树的 高度 是指根节点和叶子节点之间最长向下路径上边的数量。

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示例 1：

输入：n = 4, edges = [[1,0],[1,2],[1,3]]
输出：[1]
解释：如图所示，当根是标签为 1 的节点时，树的高度是 1，这是唯一的最小高度树。

示例 2：

输入：n = 6, edges = [[3,0],[3,1],[3,2],[3,4],[5,4]]
输出：[3,4]

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提示：

• 1 <= n <= 2 * 10^4
• edges.length == n - 1
• 0 <= ai, bi < n
• ai != bi
• 所有 (ai, bi) 互不相同
• 给定的输入 保证 是一棵树，并且 不会有重复的边

2. 🎯 s.1 - 拓扑排序剥叶

c

int* findMinHeightTrees(int n, int** edges, int edgesSize, int* edgesColSize, int* returnSize) {
    if (n == 1) {
        int* res = (int*)malloc(sizeof(int));
        res[0] = 0; *returnSize = 1;
        return res;
    }
    int* degree = (int*)calloc(n, sizeof(int));
    int** graph = (int**)malloc(sizeof(int*) * n);
    int* gSize = (int*)calloc(n, sizeof(int));
    int* gCap = (int*)malloc(sizeof(int) * n);
    for (int i = 0; i < n; i++) { gCap[i] = 4; graph[i] = (int*)malloc(sizeof(int) * 4); }
    for (int i = 0; i < edgesSize; i++) {
        int a = edges[i][0], b = edges[i][1];
        if (gSize[a] == gCap[a]) { gCap[a] *= 2; graph[a] = realloc(graph[a], sizeof(int) * gCap[a]); }
        if (gSize[b] == gCap[b]) { gCap[b] *= 2; graph[b] = realloc(graph[b], sizeof(int) * gCap[b]); }
        graph[a][gSize[a]++] = b;
        graph[b][gSize[b]++] = a;
        degree[a]++; degree[b]++;
    }
    int* leaves = (int*)malloc(sizeof(int) * n);
    int lSize = 0;
    for (int i = 0; i < n; i++) if (degree[i] == 1) leaves[lSize++] = i;
    int remaining = n;
    while (remaining > 2) {
        remaining -= lSize;
        int* newLeaves = (int*)malloc(sizeof(int) * n);
        int nlSize = 0;
        for (int i = 0; i < lSize; i++) {
            int leaf = leaves[i];
            for (int j = 0; j < gSize[leaf]; j++) {
                int nb = graph[leaf][j];
                if (--degree[nb] == 1) newLeaves[nlSize++] = nb;
            }
        }
        free(leaves);
        leaves = newLeaves;
        lSize = nlSize;
    }
    *returnSize = lSize;
    for (int i = 0; i < n; i++) free(graph[i]);
    free(graph); free(gSize); free(gCap); free(degree);
    return leaves;
}

js

/**
 * @param {number} n
 * @param {number[][]} edges
 * @return {number[]}
 */
var findMinHeightTrees = function (n, edges) {
  if (n === 1) return [0]
  const degree = new Array(n).fill(0)
  const graph = Array.from({ length: n }, () => [])
  for (const [a, b] of edges) {
    graph[a].push(b)
    graph[b].push(a)
    degree[a]++
    degree[b]++
  }
  let leaves = []
  for (let i = 0; i < n; i++) {
    if (degree[i] === 1) leaves.push(i)
  }
  let remaining = n
  while (remaining > 2) {
    remaining -= leaves.length
    const newLeaves = []
    for (const leaf of leaves) {
      for (const neighbor of graph[leaf]) {
        if (--degree[neighbor] === 1) newLeaves.push(neighbor)
      }
    }
    leaves = newLeaves
  }
  return leaves
}

py

class Solution:
    def findMinHeightTrees(self, n: int, edges: List[List[int]]) -> List[int]:
        if n == 1:
            return [0]
        degree = [0] * n
        graph = [[] for _ in range(n)]
        for a, b in edges:
            graph[a].append(b)
            graph[b].append(a)
            degree[a] += 1
            degree[b] += 1
        leaves = [i for i in range(n) if degree[i] == 1]
        remaining = n
        while remaining > 2:
            remaining -= len(leaves)
            new_leaves = []
            for leaf in leaves:
                for nb in graph[leaf]:
                    degree[nb] -= 1
                    if degree[nb] == 1:
                        new_leaves.append(nb)
            leaves = new_leaves
        return leaves

• 时间复杂度：$O(n)$，其中 $n$ 是节点数
• 空间复杂度：$O(n)$，邻接表和度数组

算法思路：

• 不断移除度为 1 的叶子节点，直到剩余节点不超过 2 个
• 最后剩下的 1 或 2 个节点即为最小高度树的根