README_EN.md

1273. Delete Tree Nodes

中文文档

Description

A tree rooted at node 0 is given as follows:

• The number of nodes is nodes;
• The value of the ith node is value[i];
• The parent of the ith node is parent[i].

Remove every subtree whose sum of values of nodes is zero.

Return the number of the remaining nodes in the tree.

 

Example 1:

Input: nodes = 7, parent = [-1,0,0,1,2,2,2], value = [1,-2,4,0,-2,-1,-1]
Output: 2

Example 2:

Input: nodes = 7, parent = [-1,0,0,1,2,2,2], value = [1,-2,4,0,-2,-1,-2]
Output: 6

 

Constraints:

• 1 <= nodes <= 104
• parent.length == nodes
• 0 <= parent[i] <= nodes - 1
• parent[0] == -1 which indicates that 0 is the root.
• value.length == nodes
• -105 <= value[i] <= 105
• The given input is guaranteed to represent a valid tree.

Solutions

Python3

class Solution:
    def deleteTreeNodes(self, nodes: int, parent: List[int], value: List[int]) -> int:
        def dfs(i):
            s, m = value[i], 1
            for j in g[i]:
                t, n = dfs(j)
                s += t
                m += n
            if s == 0:
                m = 0
            return (s, m)

        g = defaultdict(list)
        for i in range(1, nodes):
            g[parent[i]].append(i)
        return dfs(0)[1]

Java

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

    public int deleteTreeNodes(int nodes, int[] parent, int[] value) {
        g = new List[nodes];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int i = 1; i < nodes; ++i) {
            g[parent[i]].add(i);
        }
        this.value = value;
        return dfs(0)[1];
    }

    private int[] dfs(int i) {
        int[] res = new int[] {value[i], 1};
        for (int j : g[i]) {
            int[] t = dfs(j);
            res[0] += t[0];
            res[1] += t[1];
        }
        if (res[0] == 0) {
            res[1] = 0;
        }
        return res;
    }
}

C++

class Solution {
public:
    int deleteTreeNodes(int nodes, vector<int>& parent, vector<int>& value) {
        vector<vector<int>> g(nodes);
        for (int i = 1; i < nodes; ++i) {
            g[parent[i]].emplace_back(i);
        }
        function<pair<int, int>(int)> dfs = [&](int i) -> pair<int, int> {
            int s = value[i], m = 1;
            for (int j : g[i]) {
                auto [t, n] = dfs(j);
                s += t;
                m += n;
            }
            if (s == 0) {
                m = 0;
            }
            return pair<int, int>{s, m};
        };
        return dfs(0).second;
    }
};

Go

func deleteTreeNodes(nodes int, parent []int, value []int) int {
	g := make([][]int, nodes)
	for i := 1; i < nodes; i++ {
		g[parent[i]] = append(g[parent[i]], i)
	}
	type pair struct{ s, n int }
	var dfs func(int) pair
	dfs = func(i int) pair {
		s, m := value[i], 1
		for _, j := range g[i] {
			t := dfs(j)
			s += t.s
			m += t.n
		}
		if s == 0 {
			m = 0
		}
		return pair{s, m}
	}
	return dfs(0).n
}

...