Perform the given queries on the rooted tree

Difficulty Level : Hard
Last Updated : 17 Mar, 2020

Given a rooted tree and not necessarily binary. The tree contains N nodes, labeled 1 to N. You are given the tree in the form of an array A[1..N] of size N. A[i] denotes label of the parent of node labeled i. For clarity, you may assume that the tree satisfies the following conditions.

The root of the tree is labeled 1. Hence A[1] is set to 0.

The parent of node T will always have a label less than T.

The task is to perform the following operations according to the type of query given.

ADD, X, Y: add Y to the value at node X.
ADDUP, X, Y: add Y to the value at node X. Then, add Y to the value at A[X] (i.e. the parent of X). The, add Y to the value at A[A[X]] (i.e. the parent of A[X]).. and so on, till you add Y to the value at root.

After you have performed all the given operations, you are asked to answer several queries of the following type

VAL, X: print the value at node X.
VALTREE, X: print the sum of values at all nodes in the subtree rooted at X (including X).

Source: Directi Interview | Set 13

Examples:

Input:
N = 7, M = 4, Q = 5
0 1 2 2 2 1 2
ADD 6 76
ADDUP 1 49
ADD 4 48
ADDUP 2 59
VALTREE 1
VALTREE 5
VAL 5
VALTREE 2
VAL 2

Output:
291
0
0
107
59

Input:
N = 5, M = 5, Q = 3
0 1 1 1 3
ADD 1 10
ADD 2 20
ADD 3 30
ADD 4 40
ADDUP 5 50
VAL 3
VALTREE 3
VALTREE 1

Output:
80
130
250

Explanation: This problem is a slight variation of dfs. In this, we have stored the node’s original value and addup value in the vector of the pair. We did 2 times dfs.

dfs1 for offline queries i.e to calculate the addup sum for each node.
dfs2 to store the subtree sum in an array.

Now all the queries can be answered in a constant time.

Graph before dfs1

Graph after dfs1

Below is the required implementation:

C++

// C++ implementation to perform  
// above operations and queries 
#include <bits/stdc++.h> 
using namespace std; 
  
/* 
Code Parameters  
p->for every node first value is it's original value 
and second value is it's addup value 
subtree_sum[]-> to store the subtree_sum at every node 
visit-> for dfs1 
visit2->for dfs2 
*/
vector<pair<int, int> > p; 
vector<int> adj[10000]; 
int subtree_sum[10000], visit[10000], visit2[10000]; 
  
int dfs1(int root) 
{ 
    // for leaf node 
    if (adj[root].size() == 0) { 
  
        // if leaf node then add the addup 
        // sum to it's original value 
        p[root].first += p[root].second; 
        return 0; 
    } 
  
    int sum = 0; 
  
    for (int i = 0; i < adj[root].size(); i++) { 
        if (visit[adj[root][i]] == 0) { 
            dfs1(adj[root][i]); 
  
            // add the addup sum of all the adjacent 
            // neighbors to the current node 
            p[root].second += p[adj[root][i]].second; 
            visit[adj[root][i]] = 1; 
        } 
    } 
  
    // process the root node 
    p[root].first += p[root].second; 
  
    return 0; 
} 
  
int dfs2(int root) 
{ 
    if (adj[root].size() == 0) { 
  
        // for the leaf node subtree_sum 
        // will be it's own value 
        subtree_sum[root] = p[root].first; 
        return p[root].first; 
    } 
  
    int sum = p[root].first; 
  
    for (int i = 0; i < adj[root].size(); i++) { 
        if (visit2[adj[root][i]] == 0) { 
            sum += dfs2(adj[root][i]); 
            visit2[adj[root][i]] = 1; 
        } 
    } 
  
    // calculate the subtree_sum 
    // for the particular root node 
    subtree_sum[root] = sum; 
  
    return sum; 
} 
  
// Driver code 
int main() 
{ 
  
    int nodes = 7, m = 4, qu = 5, b; 
    int a[] = { 0, 1, 2, 2, 2, 1, 2 }; 
    // for root node 
    p.push_back(make_pair(0, 0)); 
  
    for (int i = 0; i < nodes; i++) { 
  
        if (a[i] != 0) 
            adj[a[i]].push_back(i + 1); 
  
        // for every node 
        p.push_back(make_pair(0, 0)); 
    } 
  
    vector<pair<string, pair<int, int> > > v; 
    v.push_back(make_pair("ADD", make_pair(6, 76))); 
    v.push_back(make_pair("ADDUP", make_pair(1, 49))); 
    v.push_back(make_pair("ADD", make_pair(4, 48))); 
    v.push_back(make_pair("ADDUP", make_pair(2, 59))); 
  
    for (int i = 0; i < m; i++) { 
        string s = v[i].first; 
        int a = v[i].second.first; 
        int b = v[i].second.second; 
        if (s == "ADD") 
            // adding to it's own value 
            p[a].first += b; 
  
        else
            // adding to it's addup value 
            p[a].second += b; 
    } 
  
    // to process the offline  queries 
    dfs1(1); 
  
    // to store the subtree sum for every root node 
    dfs2(1); 
  
    vector<pair<string, int> > q; 
    q.push_back(make_pair("VALTREE", 1)); 
    q.push_back(make_pair("VALTREE", 5)); 
    q.push_back(make_pair("VAL", 5)); 
    q.push_back(make_pair("VALTREE", 2)); 
    q.push_back(make_pair("VAL", 2)); 
    for (int i = 0; i < qu; i++) { 
        string s = q[i].first; 
        int a = q[i].second; 
  
        if (s == "VAL") 
            cout << p[a].first << "\n"; 
        else
            cout << subtree_sum[a] << "\n"; 
    } 
} 

Python3

# Python3 implementation to perform 
# above operations and queries 
p = [] 
adj = [0] * 10000
for i in range(10000): 
    adj[i] = [] 
subtree_sum, visit, visit2 = [0] * 10000, [0] * 10000, [0] * 10000
  
# Code Parameters 
# p->for every node first value is it's original value 
# and second value is it's addup value 
# subtree_sum[]-> to store the subtree_sum at every node 
# visit-> for dfs1 
# visit2->for dfs2 
def dfs1(root: int) -> int: 
  
    # for leaf node 
    if len(adj[root]) == 0: 
  
        # if leaf node then add the addup 
        # sum to it's original value 
        p[root][0] += p[root][1] 
        return 0
  
    summ = 0
    for i in range(len(adj[root])): 
        if visit[adj[root][i]] == 0: 
            dfs1(adj[root][i]) 
  
            # add the addup sum of all the adjacent 
            # neighbors to the current node 
            p[root][1] += p[adj[root][i]][1] 
            visit[adj[root][i]] = 1
  
    # process the root node 
    p[root][0] += p[root][1] 
  
    return 0
  
def dfs2(root: int) -> int: 
    if len(adj[root]) == 0: 
  
        # for the leaf node subtree_sum 
        # will be it's own value 
        subtree_sum[root] = p[root][0] 
        return p[root][0] 
  
    summ = p[root][0] 
  
    for i in range(len(adj[root])): 
        if visit2[adj[root][i]] == 0: 
            summ += dfs2(adj[root][i]) 
            visit2[adj[root][i]] = 1
  
    # calculate the subtree_sum 
    # for the particular root node 
    subtree_sum[root] = summ 
    return summ 
  
# Driver Code 
if __name__ == "__main__": 
  
    nodes, m, qu = 7, 4, 5
    a = [0, 1, 2, 2, 2, 1, 2] 
  
    # for root node 
    p.append([0, 0]) 
  
    for i in range(nodes): 
        if a[i] != 0: 
            adj[a[i]].append(i + 1) 
  
        # for every node 
        p.append([0, 0]) 
  
    v = [] 
    v.append(("ADD", [6, 76])) 
    v.append(("ADDUP", [1, 49])) 
    v.append(("ADD", [4, 48])) 
    v.append(("ADDUP", [2, 59])) 
  
    for i in range(m): 
        s = v[i][0] 
        a = v[i][1][0] 
        b = v[i][1][1] 
  
        if s == "ADD": 
  
            # adding to it's own value 
            p[a][0] += b 
        else: 
  
            # adding to it's addup value 
            p[a][1] += b 
  
    # to process the offline queries 
    dfs1(1) 
  
    # to store the subtree sum for every root node 
    dfs2(1) 
  
    q = [] 
    q.append(["VALTREE", 1]) 
    q.append(["VALTREE", 5]) 
    q.append(["VAL", 5]) 
    q.append(["VALTREE", 2]) 
    q.append(["VAL", 2]) 
    for i in range(qu): 
        s = q[i][0] 
        a = q[i][1] 
  
        if s == "VAL": 
            print(p[a][0]) 
        else: 
            print(subtree_sum[a]) 
  
# This code is contributed by 
# sanjeev2552 

Output:

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