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3249. Count the Number of Good Nodes

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int countGoodNodes(vector<vector<int>>& edges) {
    const int n = edges.size() + 1;
    int ans = 0;
    vector<vector<int>> graph(n);

    for (const vector<int>& edge : edges) {
      const int u = edge[0];
      const int v = edge[1];
      graph[u].push_back(v);
      graph[v].push_back(u);
    }

    dfs(graph, 0, /*prev=*/-1, ans);
    return ans;
  }

 private:
  int ans = 0;

  // Returns the size of the subtree rooted at u.
  int dfs(const vector<vector<int>>& graph, int u, int prev, int& ans) {
    int size = 1;
    vector<int> childrenSizes;

    for (const int v : graph[u]) {
      if (v == prev)
        continue;
      const int childSize = dfs(graph, v, u, ans);
      size += childSize;
      childrenSizes.push_back(childSize);
    }

    if (childrenSizes.empty() || allSameSizes(childrenSizes))
      ++ans;

    return size;
  }

 private:
  bool allSameSizes(const vector<int>& childrenSizes) {
    for (int i = 1; i < childrenSizes.size(); ++i)
      if (childrenSizes[i] != childrenSizes[0])
        return false;
    return true;
  }
};

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class Solution {
  public int countGoodNodes(int[][] edges) {
    final int n = edges.length + 1;
    List<Integer>[] graph = new ArrayList<>(n);

    for (int i = 0; i < n; ++i)
      graph[i] = new ArrayList<>();

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      graph[u].add(v);
      graph[v].add(u);
    }

    dfs(graph, 0, -1);
    return ans;
  }

  private int ans = 0;

  // Returns the size of the subtree rooted at u.
  private int dfs(List<Integer>[] graph, int u, int prev) {
    int size = 1;
    List<Integer> childrenSizes = new ArrayList<>();

    for (final int v : graph[u]) {
      if (v == prev)
        continue;
      final int childSize = dfs(graph, v, u);
      size += childSize;
      childrenSizes.add(childSize);
    }

    if (childrenSizes.isEmpty() || allSameSizes(childrenSizes))
      ++ans;

    return size;
  }

  private boolean allSameSizes(List<Integer> childrenSizes) {
    for (int i = 1; i < childrenSizes.size(); ++i)
      if (!childrenSizes.get(i).equals(childrenSizes.get(0)))
        return false;
    return true;
  }
}

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class Solution:
  def countGoodNodes(self, edges: list[list[int]]) -> int:
    n = len(edges) + 1
    graph = [[] for _ in range(n)]

    for u, v in edges:
      graph[u].append(v)
      graph[v].append(u)

    ans = 0

    def dfs(u: int, prev: int) -> int:
      """Returns the size of the subtree rooted at u."""
      nonlocal ans
      size = 1
      childrenSizes = []
      for v in graph[u]:
        if v == prev:
          continue
        child_size = dfs(v, u)
        size += child_size
        childrenSizes.append(child_size)

      if not childrenSizes or all(s == childrenSizes[0]
                                  for s in childrenSizes):
        ans += 1

      return size

    dfs(0, -1)
    return ans