Queries to find sum of distance of a given node to every leaf node in a Weighted Tree

Given a Undirected Weighted Tree having N nodes and E edges. Given Q queries, with each query indicating a starting node. The task is to print the sum of the distances from a given starting node S to every leaf node in the Weighted Tree.

Examples: 

Input:  N = 5, E = 4, Q = 3                    
                    1
            (4) /    \ (2)
              /        \
            4          2
                 (5)/    \ (3)
                  /        \
               5           3
Query 1: S =
Query 2: S = 3 
Query 3: S = 5
Output : 
16
17
19
Explanation : 
The three leaf nodes in the tree are 3, 4 and 5. 
For S = 1, the sum of the distances from node 1 to the leaf nodes are: d(1, 4) + d(1, 3) + d(1, 5) = 4 + (2 + 5) + (2 + 3) = 16.
For S = 3, the sum of the distances from node 3 to its leaf nodes are: d(3, 4) + d(3, 3) + d(3, 5) = (3 + 2 + 4) + 0 + (3 + 5) = 17
For S = 5, the sum of the distances from node 5 to its leaf nodes are: d(5, 4) + d(5, 3) + d(5, 5) = (5 + 2 + 4) + (5 + 3) + 0 = 19
 

Input:  N = 3, E = 2, Q = 2                    
                    1             
            (9) /    \ (1)
              /        \  
                     3  
Query 1: S =
Query 2: S = 2 
Query 3: S = 3 
Output : 
10
10
10 

Naive Approach:
For each query, traverse the entire tree and find the sum of the distance from the given source node to all the leaf nodes.
Time Complexity: O(Q * N)

Efficient Approach: The idea is to use pre-compute the sum of the distance of every node to all the leaf nodes using Dynamic Programming on trees Algorithm and obtain the answer for each query in constant time.



Follow the steps below to solve the problem:

  • Initialize a vector dp to store the sum of the distances from each node i to all the leaf nodes of the tree.
  • Initialize a vector leaves to store the count of the leaf nodes in the sub-tree of node i considering 1 as the root node.
  • Find the the sum of the distances from node i to all the leaf nodes in the sub-tree of i considering 1 as the root node using a modified Depth First Search Algorithm.

    Let node a be the parent of node i

    • leaves[a] += leaves[i] ;
    • dp[a] += dp[i] + leaves[i] * weight of edge between nodes(a, i) ;
  • Use the re-rooting technique to find the distance of the remaining leaves of the tree that are not in the sub-tree of node i. To calculate these distances, use another modified Depth First Search (DFS) algorithm to find and add the sum of the distances of leaf nodes to node i.

    Let a be the parent node and i be the child node, then
    Let the number of leaf nodes outside the sub-tree i that are present in the sub-tree a be L

    • L = leaves[a] – leaves[i] ;
    • dp[i] += ( dp[a] – dp[i] ) + ( weight of edge between nodes(a, i) ) * ( L – leaves[i] ) ;
    • leaves[i] += L ;

Below is the implementation of the above approach:

CPP

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// CPP program for the above problem
#include <bits/stdc++.h>
using namespace std;
  
// MAX size
const int N = 1e5 + 5;
  
// graph with {destination, weight};
vector<vector<pair<int, int> > > v(N);
  
// for storing the sum for ith node
vector<int> dp(N);
  
// leaves in subtree of ith.
vector<int> leaves(N);
int n;
  
// dfs to find sum of distance
// of leaves in the
// subtree of a node
void dfs(int a, int par)
{
    // flag is the node is
    // leaf or not;
    bool leaf = 1;
    for (auto& i : v[a]) {
        // skipping if parent
        if (i.first == par)
            continue;
  
        // setting flag to false
        leaf = 0;
  
        // doing dfs call
        dfs(i.first, a);
    }
  
    // doing calculation
    // in postorder.
    if (leaf == 1) {
  
        // if the node is leaf then
        // we just increment
        // the no. of leaves under
        // the subtree of a node
        leaves[a] += 1;
    }
    else {
  
        for (auto& i : v[a]) {
            if (i.first == par)
                continue;
  
            // adding num of leaves
            leaves[a]
                += leaves[i.first];
  
            // calculating answer for
            // the sum in the subtree
            dp[a] = dp[a]
                    + dp[i.first]
                    + leaves[i.first]
                          * i.second;
        }
    }
}
  
// dfs function to find the
// sum of distance of leaves
// outside the subtree
void dfs2(int a, int par)
{
    for (auto& i : v[a]) {
        if (i.first == par)
            continue;
  
        // number of leaves other
        // than the leaves in the
        // subtree of i
        int leafOutside = leaves[a] - leaves[i.first];
  
        // adding the contribution
        // of leaves outside to
        // the ith node
        dp[i.first] += (dp[a] - dp[i.first]);
  
        dp[i.first] += i.second
                       * (leafOutside
                          - leaves[i.first]);
  
        // adding the leafs outside
        // to ith node's leaves.
        leaves[i.first]
            += leafOutside;
        dfs2(i.first, a);
    }
}
  
void answerQueries(
    vector<int> queries)
{
  
    // calculating the sum of
    // distance of leaves in the
    // subtree of a node assuming
    // the root of the tree is 1
    dfs(1, 0);
  
    // calculating the sum of
    // distance of leaves outside
    // the subtree of node
    // assuming the root of the
    // tree is 1
    dfs2(1, 0);
  
    // answering the queries;
    for (int i = 0;
         i < queries.size(); i++) {
        cout << dp[queries[i]] << endl;
    }
}
  
// Driver Code
int main()
{
    // Driver Code
    /*
             1
       (4) /   \ (2)
          /     \
         4       2
             (5)/  \ (3)
               /    \
              5      3
    */
  
    n = 5;
  
    // initialising tree
    v[1].push_back(
        make_pair(4, 4));
    v[4].push_back(
        make_pair(1, 4));
    v[1].push_back(
        make_pair(2, 2));
    v[2].push_back(
        make_pair(1, 2));
    v[2].push_back(
        make_pair(3, 3));
    v[3].push_back(
        make_pair(2, 3));
    v[2].push_back(
        make_pair(5, 5));
    v[5].push_back(
        make_pair(2, 5));
  
    vector<int>
        queries = { 1, 3, 5 };
    answerQueries(queries);
}

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Output:

16
17
19

Time Complexity: O(N + Q)
Auxiliary Space: O(N)

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