Sum of lengths of all paths possible in a given tree

Given a tree with N nodes, the task is to find the sum of lengths of all the paths. Path length for two nodes in the tree is the number of edges on the path and for two adjacent nodes in the tree, the length of the path is 1.

Examples:

Input:
       0
     /   \
   1      2
 /  \
3    4 
Output: 18
Paths of length 1 = (0, 1), (0, 2), (1, 3), (1, 4) = 4
Paths of length 2 = (0, 3), (0, 4), (1, 2), (3, 4) = 4
Paths of length 3 = (3, 2), (4, 2) = 2
The sum of lengths of all paths = 
    (4 * 1) + (4 * 2) + (2 * 3) = 18

Input:
       0
     /
   1
 /
2
Output: 4

Naive approach: Check all possible paths and then add them to compute the final result. The complexity of this approach will be O(n2).



Efficient approach: It can be noted that each edge in a tree is a bridge. Hence that edge is going to be present in every path possible between the two subtrees that the edge connects.
For example, the edge (1 – 0) is present in every path possible between {1, 3, 4} and {0, 2}, (1 – 0) is used for 6 times that is size of the subtree {1, 3, 4} multiplied by the size of the subtree {0, 2}. So for each edge, compute how many times that edge is going to be considered for the paths going over it. DFS can be used to store the size of the subtree and the contribution of all edges can be computed with another dfs. The complexity of this approach will be O(n).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int sz = 1e5;
  
// Number of vertices
int n;
  
// Adjacency list representation
// of the tree
vector<int> tree[sz];
  
// Array that stores the subtree size
int subtree_size[sz];
  
// Array to mark all the
// vertices which are visited
int vis[sz];
  
// Utility function to create an
// edge between two vertices
void addEdge(int a, int b)
{
  
    // Add a to b's list
    tree[a].push_back(b);
  
    // Add b to a's list
    tree[b].push_back(a);
}
  
// Function to calculate the subtree size
int dfs(int node)
{
  
    // Mark visited
    vis[node] = 1;
    subtree_size[node] = 1;
  
    // For every adjacent node
    for (auto child : tree[node]) {
  
        // If not already visited
        if (!vis[child]) {
  
            // Recursive call for the child
            subtree_size[node] += dfs(child);
        }
    }
    return subtree_size[node];
}
  
// Function to calculate the
// contribution of each edge
void contribution(int node, int& ans)
{
  
    // Mark current node as visited
    vis[node] = true;
  
    // For every adjacent node
    for (int child : tree[node]) {
  
        // If it is not already visisted
        if (!vis[child]) {
            ans += (subtree_size[child]
                    * (n - subtree_size[child]));
            contribution(child, ans);
        }
    }
}
  
// Function to return the required sum
int getSum()
{
  
    // First pass of the dfs
    memset(vis, 0, sizeof(vis));
    dfs(0);
  
    // Second pass
    int ans = 0;
    memset(vis, 0, sizeof(vis));
    contribution(0, ans);
  
    return ans;
}
  
// Driver code
int main()
{
    n = 5;
  
    addEdge(0, 1);
    addEdge(0, 2);
    addEdge(1, 3);
    addEdge(1, 4);
  
    cout << getSum();
  
    return 0;
}

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Python3

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# Python3 implementation of the approach
sz = 10**5
  
# Number of vertices
n = 5
an = 0
  
# Adjacency list representation
# of the tree
tree = [[] for i in range(sz)]
  
# Array that stores the subtree size
subtree_size = [0] * sz
  
# Array to mark all the
# vertices which are visited
vis = [0] * sz
  
# Utility function to create an
# edge between two vertices
def addEdge(a, b):
  
    # Add a to b's list
    tree[a].append(b)
  
    # Add b to a's list
    tree[b].append(a)
  
# Function to calculate the subtree size
def dfs(node):
    leaf = True
  
    # Mark visited
    vis[node] = 1
      
    # For every adjacent node
    for child in tree[node]:
  
        # If not already visited
        if (vis[child] == 0):
            leaf = False
            dfs(child)
              
            # Recursive call for the child
            subtree_size[node] += subtree_size[child]
  
    if leaf:
        subtree_size[node] = 1
  
# Function to calculate the
# contribution of each edge
def contribution(node,ans):
    global an
  
    # Mark current node as visited
    vis[node] = 1
  
    # For every adjacent node
    for child in tree[node]:
  
        # If it is not already visisted
        if (vis[child] == 0):
            an += (subtree_size[child] * 
              (n - subtree_size[child]))
            contribution(child, ans)
  
# Function to return the required sum
def getSum():
  
    # First pass of the dfs
    for i in range(sz):
        vis[i] = 0
    dfs(0)
  
    # Second pass
    ans = 0
    for i in range(sz):
        vis[i] = 0
  
    contribution(0, ans)
  
    return an
  
# Driver code
n = 5
  
addEdge(0, 1)
addEdge(0, 2)
addEdge(1, 3)
addEdge(1, 4)
  
print(getSum())
  
# This code is contributed by Mohit Kumar

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Output:

18


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