1273. Delete Tree Nodes

1273. Delete Tree Nodes (Medium)

A tree rooted at node 0 is given as follows:

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

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

After doing so, return the number of nodes remaining 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

Example 3:

Input: nodes = 5, parent = [-1,0,1,0,0], value = [-672,441,18,728,378]
Output: 5

Example 4:

Input: nodes = 5, parent = [-1,0,0,1,1], value = [-686,-842,616,-739,-746]
Output: 5

 

Constraints:

• 1 <= nodes <= 10^4
• parent.length == nodes
• 0 <= parent[i] <= nodes - 1
• parent[0] == -1 which indicates that 0 is the root.
• value.length == nodes
• -10^5 <= value[i] <= 10^5
• The given input is guaranteed to represent a valid tree.

Companies:
Microsoft

Related Topics:
Tree, Depth-First Search, Breadth-First Search

Solution 1. DFS

// OJ: https://leetcode.com/problems/delete-tree-nodes/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
    vector<vector<int>> G;
    pair<int, int> dfs(int u, vector<int> &value) {
        int sum = value[u], cnt = 1;
        for (int v : G[u]) {
            auto [c, s] = dfs(v, value);
            sum += s;
            cnt += c;
        }
        if (sum == 0) cnt = 0;
        return { cnt, sum };
    }
public:
    int deleteTreeNodes(int nodes, vector<int>& parent, vector<int>& value) {
        G.assign(nodes, {});
        for (int i = 1; i < nodes; ++i) {
            G[parent[i]].push_back(i);
        }
        return dfs(0, value).first;
    }
};