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Weekly Contest 410 Q2
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3249. Count the Number of Good Nodes

中文文档

Description

There is an undirected tree with n nodes labeled from 0 to n - 1, and rooted at node 0. You are given a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.

A node is good if all the subtrees rooted at its children have the same size.

Return the number of good nodes in the given tree.

A subtree of treeName is a tree consisting of a node in treeName and all of its descendants.

 

Example 1:

Input: edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]

Output: 7

Explanation:

All of the nodes of the given tree are good.

Example 2:

Input: edges = [[0,1],[1,2],[2,3],[3,4],[0,5],[1,6],[2,7],[3,8]]

Output: 6

Explanation:

There are 6 good nodes in the given tree. They are colored in the image above.

Example 3:

Input: edges = [[0,1],[1,2],[1,3],[1,4],[0,5],[5,6],[6,7],[7,8],[0,9],[9,10],[9,12],[10,11]]

Output: 12

Explanation:

All nodes except node 9 are good.

 

Constraints:

  • 2 <= n <= 105
  • edges.length == n - 1
  • edges[i].length == 2
  • 0 <= ai, bi < n
  • The input is generated such that edges represents a valid tree.

Solutions

Solution 1: DFS

First, we construct the adjacency list $\textit{g}$ of the tree based on the given edges $\textit{edges}$, where $\textit{g}[a]$ represents all the neighboring nodes of node $a$.

Next, we design a function $\textit{dfs}(a, \textit{fa})$ to calculate the number of nodes in the subtree rooted at node $a$ and to accumulate the count of good nodes. Here, $\textit{fa}$ represents the parent node of node $a$.

The execution process of the function $\textit{dfs}(a, \textit{fa})$ is as follows:

  1. Initialize variables $\textit{pre} = -1$, $\textit{cnt} = 1$, $\textit{ok} = 1$, representing the number of nodes in a subtree of node $a$, the total number of nodes in all subtrees of node $a$, and whether node $a$ is a good node, respectively.
  2. Traverse all neighboring nodes $b$ of node $a$. If $b$ is not equal to $\textit{fa}$, recursively call $\textit{dfs}(b, a)$, with the return value being $\textit{cur}$, and add $\textit{cur}$ to $\textit{cnt}$. If $\textit{pre} &lt; 0$, assign $\textit{cur}$ to $\textit{pre}$; otherwise, if $\textit{pre}$ is not equal to $\textit{cur}$, it means the number of nodes in different subtrees of node $a$ is different, and set $\textit{ok}$ to $0$.
  3. Finally, add $\textit{ok}$ to the answer and return $\textit{cnt}$.

In the main function, we call $\textit{dfs}(0, -1)$ and return the final answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ represents the number of nodes.

Python3

class Solution:
    def countGoodNodes(self, edges: List[List[int]]) -> int:
        def dfs(a: int, fa: int) -> int:
            pre = -1
            cnt = ok = 1
            for b in g[a]:
                if b != fa:
                    cur = dfs(b, a)
                    cnt += cur
                    if pre < 0:
                        pre = cur
                    elif pre != cur:
                        ok = 0
            nonlocal ans
            ans += ok
            return cnt

        g = defaultdict(list)
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        ans = 0
        dfs(0, -1)
        return ans

Java

class Solution {
    private int ans;
    private List<Integer>[] g;

    public int countGoodNodes(int[][] edges) {
        int n = edges.length + 1;
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        dfs(0, -1);
        return ans;
    }

    private int dfs(int a, int fa) {
        int pre = -1, cnt = 1, ok = 1;
        for (int b : g[a]) {
            if (b != fa) {
                int cur = dfs(b, a);
                cnt += cur;
                if (pre < 0) {
                    pre = cur;
                } else if (pre != cur) {
                    ok = 0;
                }
            }
        }
        ans += ok;
        return cnt;
    }
}

C++

class Solution {
public:
    int countGoodNodes(vector<vector<int>>& edges) {
        int n = edges.size() + 1;
        vector<int> g[n];
        for (const auto& e : edges) {
            int a = e[0], b = e[1];
            g[a].push_back(b);
            g[b].push_back(a);
        }
        int ans = 0;
        auto dfs = [&](this auto&& dfs, int a, int fa) -> int {
            int pre = -1, cnt = 1, ok = 1;
            for (int b : g[a]) {
                if (b != fa) {
                    int cur = dfs(b, a);
                    cnt += cur;
                    if (pre < 0) {
                        pre = cur;
                    } else if (pre != cur) {
                        ok = 0;
                    }
                }
            }
            ans += ok;
            return cnt;
        };
        dfs(0, -1);
        return ans;
    }
};

Go

func countGoodNodes(edges [][]int) (ans int) {
	n := len(edges) + 1
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	var dfs func(int, int) int
	dfs = func(a, fa int) int {
		pre, cnt, ok := -1, 1, 1
		for _, b := range g[a] {
			if b != fa {
				cur := dfs(b, a)
				cnt += cur
				if pre < 0 {
					pre = cur
				} else if pre != cur {
					ok = 0
				}
			}
		}
		ans += ok
		return cnt
	}
	dfs(0, -1)
	return
}

TypeScript

function countGoodNodes(edges: number[][]): number {
    const n = edges.length + 1;
    const g: number[][] = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    let ans = 0;
    const dfs = (a: number, fa: number): number => {
        let [pre, cnt, ok] = [-1, 1, 1];
        for (const b of g[a]) {
            if (b !== fa) {
                const cur = dfs(b, a);
                cnt += cur;
                if (pre < 0) {
                    pre = cur;
                } else if (pre !== cur) {
                    ok = 0;
                }
            }
        }
        ans += ok;
        return cnt;
    };
    dfs(0, -1);
    return ans;
}