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

  • Last Updated : 18 Jun, 2021

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 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:

C++




// C++ 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);
}

Java




// Java program for the above problem
import java.util.*;
import java.lang.*;
 
class GFG{
     
static class pair
{
    int first, second;
     
    pair(int f, int s)
    {
        this.first = f;
        this.second = s;
    }
}
 
// MAX size
static final int N = (int)1e5 + 5;
  
// Graph with {destination, weight};
static ArrayList<ArrayList<pair>> v;
  
// For storing the sum for ith node
static int[] dp = new int[N];
  
// Leaves in subtree of ith.
static int[] leaves = new int[N];
static int n;
  
// dfs to find sum of distance
// of leaves in the subtree of
// a node
static void dfs(int a, int par)
{
     
    // Flag is the node is
    // leaf or not;
    int leaf = 1;
     
    for(pair i : v.get(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(pair i : v.get(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
static void dfs2(int a, int par)
{
    for(pair i : v.get(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);
    }
}
  
static void answerQueries(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.length; i++)
    {
        System.out.println(dp[queries[i]]);
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    /*
             1
       (4) /   \ (2)
          /     \
         4       2
             (5)/  \ (3)
               /    \
              5      3
    */
     
    n = 5;
     
    v = new ArrayList<>();
    for(int i = 0; i <= n; i++)
        v.add(new ArrayList<pair>());
     
    // Initialising tree
    v.get(1).add(new pair(4, 4));
    v.get(4).add(new pair(1, 4));
    v.get(1).add(new pair(2, 2));
    v.get(2).add(new pair(1, 2));
    v.get(2).add(new pair(3, 3));
    v.get(3).add(new pair(2, 3));
    v.get(2).add(new pair(5, 5));
    v.get(5).add(new pair(2, 5));
     
    // System.out.println(v);
    int[] queries = { 1, 3, 5 };
     
    answerQueries(queries);
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program for the above problem
 
# MAX size
N = 10 ** 5 + 5
 
# Graph with {destination, weight}
v = [[] for i in range(N)]
 
# For storing the sum for ith node
dp = [0] * N
 
# Leaves in subtree of ith.
leaves = [0] * (N)
n = 0
 
# dfs to find sum of distance
# of leaves in the subtree of
# a node
def dfs(a, par):
     
    # flag is the node is
    # leaf or not
    leaf = 1
     
    for i in v[a]:
         
        # Skipping if parent
        if (i[0] == par):
            continue
 
        # Setting flag to false
        leaf = 0
 
        # Doing dfs call
        dfs(i[0], 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 i in v[a]:
            if (i[0] == par):
                continue
 
            # Adding num of leaves
            leaves[a] += leaves[i[0]]
 
            # Calculating answer for
            # the sum in the subtree
            dp[a] = (dp[a] + dp[i[0]] +
                 leaves[i[0]] * i[1])
 
# dfs function to find the
# sum of distance of leaves
# outside the subtree
def dfs2(a, par):
     
    for i in v[a]:
        if (i[0] == par):
            continue
 
        # Number of leaves other
        # than the leaves in the
        # subtree of i
        leafOutside = leaves[a] - leaves[i[0]]
 
        # Adding the contribution
        # of leaves outside to
        # the ith node
        dp[i[0]] += (dp[a] - dp[i[0]])
 
        dp[i[0]] += i[1] * (leafOutside -
                            leaves[i[0]])
 
        # Adding the leafs outside
        # to ith node's leaves.
        leaves[i[0]] += leafOutside
        dfs2(i[0], a)
 
def answerQueries(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 i in range(len(queries)):
        print(dp[queries[i]])
 
# Driver Code
if __name__ == '__main__':
     
    #          1
    #    (4) /   \ (2)
    #       /     \
    #      4       2
    #          (5)/  \ (3)
    #            /    \
    #           5      3
    #
 
    n = 5
 
    # Initialising tree
    v[1].append([4, 4])
    v[4].append([1, 4])
    v[1].append([2, 2])
    v[2].append([1, 2])
    v[2].append([3, 3])
    v[3].append([2, 3])
    v[2].append([5, 5])
    v[5].append([2, 5])
 
    queries = [ 1, 3, 5 ]
     
    answerQueries(queries)
 
# This code is contributed by mohit kumar 29

Javascript




<script>
 
// JavaScript program for the above problem
 
class pair
{
    constructor(f,s)
    {
        this.first = f;
        this.second = s;
    }
}
 
// MAX size
let N = 1e5 + 5;
 
// Graph with {destination, weight};
let v=[];
 
// For storing the sum for ith node
let dp = new Array(N);
 
 
// Leaves in subtree of ith.
let leaves = new Array(N);
for(let i=0;i<N;i++)
{
    dp[i]=0;
    leaves[i]=0;
}
 
let n;
 
// dfs to find sum of distance
// of leaves in the subtree of
// a node
function dfs(a,par)
{
    // Flag is the node is
    // leaf or not;
    let leaf = 1;
      
    for(let i=0;i<v[a].length;i++)
    {
          
        // Skipping if parent
        if (v[a][i].first == par)
            continue;
   
        // Setting flag to false
        leaf = 0;
   
        // Doing dfs call
        dfs(v[a][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(let i=0;i<v[a].length;i++)
        {
            if (v[a][i].first == par)
                continue;
   
            // Adding num of leaves
            leaves[a] += leaves[v[a][i].first];
   
            // Calculating answer for
            // the sum in the subtree
            dp[a] = dp[a] + dp[v[a][i].first] +
                        leaves[v[a][i].first] *
                               v[a][i].second;
        }
    }
}
 
// dfs function to find the
// sum of distance of leaves
// outside the subtree
function dfs2(a,par)
{
    for(let i=0;i< v[a].length;i++)
    {
        if (v[a][i].first == par)
            continue;
   
        // Number of leaves other
        // than the leaves in the
        // subtree of i
        let leafOutside = leaves[a] -
                          leaves[v[a][i].first];
   
        // Adding the contribution
        // of leaves outside to
        // the ith node
        dp[v[a][i].first] += (dp[a] - dp[v[a][i].first]);
   
        dp[v[a][i].first] += v[a][i].second *
                   (leafOutside -
                    leaves[v[a][i].first]);
   
        // Adding the leafs outside
        // to ith node's leaves.
        leaves[v[a][i].first] += leafOutside;
        dfs2(v[a][i].first, a);
    }
}
 
function answerQueries(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(let i = 0; i < queries.length; i++)
    {
        document.write(dp[queries[i]]+"<br>");
    }
}
 
// Driver code
/*
             1
       (4) /   \ (2)
          /     \
         4       2
             (5)/  \ (3)
               /    \
              5      3
    */
 
n = 5;
 
v = [];
for(let i = 0; i <= n; i++)
    v.push([]);
 
// Initialising tree
v[1].push(new pair(4, 4));
v[4].push(new pair(1, 4));
v[1].push(new pair(2, 2));
v[2].push(new pair(1, 2));
v[2].push(new pair(3, 3));
v[3].push(new pair(2, 3));
v[2].push(new pair(5, 5));
v[5].push(new pair(2, 5));
 
// System.out.println(v);
let queries = [ 1, 3, 5 ];
 
answerQueries(queries);
 
 
// This code is contributed by patel2127
 
</script>
Output: 
16
17
19

 

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

 

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