Given a binary tree having N nodes and weight of N-1 edges. The distance between two nodes is the sum of the weight of edges on the path between two nodes. Each query contains two integers U and V, the task is to find the distance between nodes U and V.
Examples:
Input:
Output: 3 5 12 12
Explanation:
Distance between nodes 1 to 3 = weight(1, 3) = 2
Distance between nodes 2 to 3 = weight(1, 2) + weight(1, 3) = 5
Distance between nodes 3 to 5 = weight(1, 3) + weight(1, 2) + weight(2, 5) = 12
Distance between nodes 4 to 5 = weight(4, 2) + weight(2, 5) = 12
Approach: The idea is to use LCA in a tree using Binary Lifting Technique.
- Binary Lifting is a Dynamic Programming approach where we pre-compute an array lca[i][j] where i = [1, n], j = [1, log(n)] and lca[i][j] contains 2j-th ancestor of node i.
- For computing the values of lca[][], the following recursion may be used
- As we will compute the lca[][] array we will also calculate the distance[][] where distance[i][j] contains the distance from node i to its 2j-th ancestor
- For computing the values of dist[][], the following recursion may be used.
- After precomputation, we find the distance between (u, v) as we find the least common ancestor of (u, v).
Below is the implementation of the above approach:
// C++ Program to find distance
// between two nodes using LCA
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000
#define log 10 // log2(MAX)
// Array to store the level
// of each node
int level[MAX];
int lca[MAX][log];
int dist[MAX][log];
// Vector to store tree
vector<pair<int, int> > graph[MAX];
void addEdge(int u, int v, int cost)
{
graph[u].push_back({ v, cost });
graph[v].push_back({ u, cost });
}
// Pre-Processing to calculate
// values of lca[][], dist[][]
void dfs(int node, int parent,
int h, int cost)
{
// Using recursion formula to
// calculate the values
// of lca[][]
lca[node][0] = parent;
// Storing the level of
// each node
level[node] = h;
if (parent != -1) {
dist[node][0] = cost;
}
for (int i = 1; i < log; i++) {
if (lca[node][i - 1] != -1) {
// Using recursion formula to
// calculate the values of
// lca[][] and dist[][]
lca[node][i]
= lca[lca[node]
[i - 1]]
[i - 1];
dist[node][i]
= dist[node][i - 1]
+ dist[lca[node][i - 1]]
[i - 1];
}
}
for (auto i : graph[node]) {
if (i.first == parent)
continue;
dfs(i.first, node,
h + 1, i.second);
}
}
// Function to find the distance
// between given nodes u and v
void findDistance(int u, int v)
{
int ans = 0;
// The node which is present
// farthest from the root node
// is taken as v. If u is
// farther from root node
// then swap the two
if (level[u] > level[v])
swap(u, v);
// Finding the ancestor of v
// which is at same level as u
for (int i = log - 1; i >= 0; i--) {
if (lca[v][i] != -1
&& level[lca[v][i]]
>= level[u]) {
// Adding distance of node
// v till its 2^i-th ancestor
ans += dist[v][i];
v = lca[v][i];
}
}
// If u is the ancestor of v
// then u is the LCA of u and v
if (v == u) {
cout << ans << endl;
}
else {
// Finding the node closest to the
// root which is not the common
// ancestor of u and v i.e. a node
// x such that x is not the common
// ancestor of u and v but lca[x][0] is
for (int i = log - 1; i >= 0; i--) {
if (lca[v][i] != lca[u][i]) {
// Adding the distance
// of v and u to
// its 2^i-th ancestor
ans += dist[u][i] + dist[v][i];
v = lca[v][i];
u = lca[u][i];
}
}
// Adding the distance of u and v
// to its first ancestor
ans += dist[u][0] + dist[v][0];
cout << ans << endl;
}
}
// Driver Code
int main()
{
// Number of nodes
int n = 5;
// Add edges with their cost
addEdge(1, 2, 2);
addEdge(1, 3, 3);
addEdge(2, 4, 5);
addEdge(2, 5, 7);
// Initialising lca and dist values
// with -1 and 0 respectively
for (int i = 1; i <= n; i++) {
for (int j = 0; j < log; j++) {
lca[i][j] = -1;
dist[i][j] = 0;
}
}
// Perform DFS
dfs(1, -1, 0, 0);
// Query 1: {1, 3}
findDistance(1, 3);
// Query 2: {2, 3}
findDistance(2, 3);
// Query 3: {3, 5}
findDistance(3, 5);
return 0;
}
// Java program to find distance
// between two nodes using LCA
import java.io.*;
import java.util.*;
class GFG{
static final int MAX = 1000;
// log2(MAX)
static final int log = 10;
// Array to store the level
// of each node
static int[] level = new int[MAX];
static int[][] lca = new int[MAX][log];
static int[][] dist = new int[MAX][log];
// Vector to store tree
@SuppressWarnings("unchecked")
static List<List<int[]> > graph = new ArrayList();
static void addEdge(int u, int v, int cost)
{
graph.get(u).add(new int[]{ v, cost });
graph.get(v).add(new int[]{ u, cost });
}
// Pre-Processing to calculate
// values of lca[][], dist[][]
static void dfs(int node, int parent,
int h, int cost)
{
// Using recursion formula to
// calculate the values
// of lca[][]
lca[node][0] = parent;
// Storing the level of
// each node
level[node] = h;
if (parent != -1)
{
dist[node][0] = cost;
}
for(int i = 1; i < log; i++)
{
if (lca[node][i - 1] != -1)
{
// Using recursion formula to
// calculate the values of
// lca[][] and dist[][]
lca[node][i] = lca[lca[node][i - 1]][i - 1];
dist[node][i] = dist[node][i - 1] +
dist[lca[node][i - 1]][i - 1];
}
}
for(int[] i : graph.get(node))
{
if (i[0] == parent)
continue;
dfs(i[0], node, h + 1, i[1]);
}
}
// Function to find the distance
// between given nodes u and v
static void findDistance(int u, int v)
{
int ans = 0;
// The node which is present
// farthest from the root node
// is taken as v. If u is
// farther from root node
// then swap the two
if (level[u] > level[v])
{
int temp = u;
u = v;
v = temp;
}
// Finding the ancestor of v
// which is at same level as u
for(int i = log - 1; i >= 0; i--)
{
if (lca[v][i] != -1 &&
level[lca[v][i]] >= level[u])
{
// Adding distance of node
// v till its 2^i-th ancestor
ans += dist[v][i];
v = lca[v][i];
}
}
// If u is the ancestor of v
// then u is the LCA of u and v
if (v == u)
{
System.out.println(ans);
}
else
{
// Finding the node closest to the
// root which is not the common
// ancestor of u and v i.e. a node
// x such that x is not the common
// ancestor of u and v but lca[x][0] is
for(int i = log - 1; i >= 0; i--)
{
if (lca[v][i] != lca[u][i])
{
// Adding the distance
// of v and u to
// its 2^i-th ancestor
ans += dist[u][i] + dist[v][i];
v = lca[v][i];
u = lca[u][i];
}
}
// Adding the distance of u and v
// to its first ancestor
ans += dist[u][0] + dist[v][0];
System.out.println(ans);
}
}
// Driver Code
public static void main(String[] args)
{
// Number of nodes
int n = 5;
for(int i = 0; i < MAX; i++)
{
graph.add(new ArrayList<int[]>());
}
// Add edges with their cost
addEdge(1, 2, 2);
addEdge(1, 3, 3);
addEdge(2, 4, 5);
addEdge(2, 5, 7);
// Initialising lca and dist values
// with -1 and 0 respectively
for(int i = 1; i <= n; i++)
{
for(int j = 0; j < log; j++)
{
lca[i][j] = -1;
dist[i][j] = 0;
}
}
// Perform DFS
dfs(1, -1, 0, 0);
// Query 1: {1, 3}
findDistance(1, 3);
// Query 2: {2, 3}
findDistance(2, 3);
// Query 3: {3, 5}
findDistance(3, 5);
}
}
// This code is contributed by jithin
# Python3 Program to find
# distance between two nodes
# using LCA
MAX = 1000
# lg2(MAX)
lg = 10
# Array to store the level
# of each node
level = [0 for i in range(MAX)]
lca = [[0 for i in range(lg)]
for j in range(MAX)]
dist = [[0 for i in range(lg)]
for j in range(MAX)]
# Vector to store tree
graph = [[] for i in range(MAX)]
def addEdge(u, v, cost):
global graph
graph[u].append([v, cost])
graph[v].append([u, cost])
# Pre-Processing to calculate
# values of lca[][], dist[][]
def dfs(node, parent, h, cost):
# Using recursion formula to
# calculate the values
# of lca[][]
lca[node][0] = parent
# Storing the level of
# each node
level[node] = h
if (parent != -1):
dist[node][0] = cost
for i in range(1, lg):
if (lca[node][i - 1] != -1):
# Using recursion formula to
# calculate the values of
# lca[][] and dist[][]
lca[node][i] = lca[lca[node][i - 1]][i - 1]
dist[node][i] = (dist[node][i - 1] +
dist[lca[node][i - 1]][i - 1])
for i in graph[node]:
if (i[0] == parent):
continue
dfs(i[0], node, h + 1, i[1])
# Function to find the distance
# between given nodes u and v
def findDistance(u, v):
ans = 0
# The node which is present
# farthest from the root node
# is taken as v. If u is
# farther from root node
# then swap the two
if (level[u] > level[v]):
temp = u
u = v
v = temp
# Finding the ancestor of v
# which is at same level as u
i = lg - 1
while(i >= 0):
if (lca[v][i] != -1 and
level[lca[v][i]] >= level[u]):
# Adding distance of node
# v till its 2^i-th ancestor
ans += dist[v][i]
v = lca[v][i]
i -= 1
# If u is the ancestor of v
# then u is the LCA of u and v
if (v == u):
print(ans)
else:
# Finding the node closest to the
# root which is not the common
# ancestor of u and v i.e. a node
# x such that x is not the common
# ancestor of u and v but lca[x][0] is
i = lg - 1
while(i >= 0):
if (lca[v][i] != lca[u][i]):
# Adding the distance
# of v and u to
# its 2^i-th ancestor
ans += dist[u][i] + dist[v][i]
v = lca[v][i]
u = lca[u][i]
i -= 1
# Adding the distance of u and v
# to its first ancestor
ans += (dist[u][0] +
dist[v][0])
print(ans)
# Driver Code
if __name__ == '__main__':
# Number of nodes
n = 5
# Add edges with their cost
addEdge(1, 2, 2)
addEdge(1, 3, 3)
addEdge(2, 4, 5)
addEdge(2, 5, 7)
# Initialising lca and dist values
# with -1 and 0 respectively
for i in range(1, n + 1):
for j in range(lg):
lca[i][j] = -1
dist[i][j] = 0
# Perform DFS
dfs(1, -1, 0, 0)
# Query 1: {1, 3}
findDistance(1, 3)
# Query 2: {2, 3}
findDistance(2, 3)
# Query 3: {3, 5}
findDistance(3, 5)
# This code is contributed by SURENDRA_GANGWAR
// C# program to find distance
// between two nodes using LCA
using System;
using System.Collections.Generic;
class GFG
{
static readonly int MAX = 1000;
// log2(MAX)
static readonly int log = 10;
// Array to store the level
// of each node
static int[] level = new int[MAX];
static int[,] lca = new int[MAX,log];
static int[,] dist = new int[MAX,log];
// List to store tree
static List<List<int[]> > graph = new List<List<int[]>>();
static void addEdge(int u, int v, int cost)
{
graph[u].Add(new int[]{ v, cost });
graph[v].Add(new int[]{ u, cost });
}
// Pre-Processing to calculate
// values of lca[,], dist[,]
static void dfs(int node, int parent,
int h, int cost)
{
// Using recursion formula to
// calculate the values
// of lca[,]
lca[node, 0] = parent;
// Storing the level of
// each node
level[node] = h;
if (parent != -1)
{
dist[node, 0] = cost;
}
for(int i = 1; i < log; i++)
{
if (lca[node, i - 1] != -1)
{
// Using recursion formula to
// calculate the values of
// lca[,] and dist[,]
lca[node, i] = lca[lca[node, i - 1], i - 1];
dist[node, i] = dist[node, i - 1] +
dist[lca[node, i - 1], i - 1];
}
}
foreach(int[] i in graph[node])
{
if (i[0] == parent)
continue;
dfs(i[0], node, h + 1, i[1]);
}
}
// Function to find the distance
// between given nodes u and v
static void findDistance(int u, int v)
{
int ans = 0;
// The node which is present
// farthest from the root node
// is taken as v. If u is
// farther from root node
// then swap the two
if (level[u] > level[v])
{
int temp = u;
u = v;
v = temp;
}
// Finding the ancestor of v
// which is at same level as u
for(int i = log - 1; i >= 0; i--)
{
if (lca[v, i] != -1 &&
level[lca[v, i]] >= level[u])
{
// Adding distance of node
// v till its 2^i-th ancestor
ans += dist[v, i];
v = lca[v, i];
}
}
// If u is the ancestor of v
// then u is the LCA of u and v
if (v == u)
{
Console.WriteLine(ans);
}
else
{
// Finding the node closest to the
// root which is not the common
// ancestor of u and v i.e. a node
// x such that x is not the common
// ancestor of u and v but lca[x,0] is
for(int i = log - 1; i >= 0; i--)
{
if (lca[v, i] != lca[u, i])
{
// Adding the distance
// of v and u to
// its 2^i-th ancestor
ans += dist[u, i] + dist[v, i];
v = lca[v, i];
u = lca[u, i];
}
}
// Adding the distance of u and v
// to its first ancestor
ans += dist[u, 0] + dist[v, 0];
Console.WriteLine(ans);
}
}
// Driver Code
public static void Main(String[] args)
{
// Number of nodes
int n = 5;
for(int i = 0; i < MAX; i++)
{
graph.Add(new List<int[]>());
}
// Add edges with their cost
addEdge(1, 2, 2);
addEdge(1, 3, 3);
addEdge(2, 4, 5);
addEdge(2, 5, 7);
// Initialising lca and dist values
// with -1 and 0 respectively
for(int i = 1; i <= n; i++)
{
for(int j = 0; j < log; j++)
{
lca[i, j] = -1;
dist[i, j] = 0;
}
}
// Perform DFS
dfs(1, -1, 0, 0);
// Query 1: {1, 3}
findDistance(1, 3);
// Query 2: {2, 3}
findDistance(2, 3);
// Query 3: {3, 5}
findDistance(3, 5);
}
}
// This code is contributed by aashish1995
<script>
// Javascript program to find distance
// between two nodes using LCA
let MAX = 1000;
// log2(MAX)
let log = 10;
// Array to store the level
// of each node
let level = new Array(MAX);
let lca = new Array(MAX);
let dist = new Array(MAX);
// Vector to store tree
let graph = [];
function addEdge(u, v, cost)
{
graph[u].push([ v, cost ]);
graph[v].push([ u, cost ]);
}
// Pre-Processing to calculate
// values of lca[][], dist[][]
function dfs(node, parent, h, cost)
{
// Using recursion formula to
// calculate the values
// of lca[][]
lca[node][0] = parent;
// Storing the level of
// each node
level[node] = h;
if (parent != -1)
{
dist[node][0] = cost;
}
for(let i = 1; i < log; i++)
{
if (lca[node][i - 1] != -1)
{
// Using recursion formula to
// calculate the values of
// lca[][] and dist[][]
lca[node][i] = lca[lca[node][i - 1]][i - 1];
dist[node][i] = dist[node][i - 1] +
dist[lca[node][i - 1]][i - 1];
}
}
for(let i = 0; i < graph[node].length; i++)
{
if (graph[node][i][0] == parent)
continue;
dfs(graph[node][i][0], node, h + 1, graph[node][i][1]);
}
}
// Function to find the distance
// between given nodes u and v
function findDistance(u, v)
{
let ans = 0;
// The node which is present
// farthest from the root node
// is taken as v. If u is
// farther from root node
// then swap the two
if (level[u] > level[v])
{
let temp = u;
u = v;
v = temp;
}
// Finding the ancestor of v
// which is at same level as u
for(let i = log - 1; i >= 0; i--)
{
if (lca[v][i] != -1 && level[lca[v][i]] >= level[u])
{
// Adding distance of node
// v till its 2^i-th ancestor
ans += dist[v][i];
v = lca[v][i];
}
}
// If u is the ancestor of v
// then u is the LCA of u and v
if (v == u)
{
document.write(ans + "</br>");
}
else
{
// Finding the node closest to the
// root which is not the common
// ancestor of u and v i.e. a node
// x such that x is not the common
// ancestor of u and v but lca[x][0] is
for(let i = log - 1; i >= 0; i--)
{
if (lca[v][i] != lca[u][i])
{
// Adding the distance
// of v and u to
// its 2^i-th ancestor
ans += dist[u][i] + dist[v][i];
v = lca[v][i];
u = lca[u][i];
}
}
// Adding the distance of u and v
// to its first ancestor
ans += dist[u][0] + dist[v][0];
document.write(ans + "</br>");
}
}
// Number of nodes
let n = 5;
for(let i = 0; i < MAX; i++)
{
graph.push([]);
}
// Add edges with their cost
addEdge(1, 2, 2);
addEdge(1, 3, 3);
addEdge(2, 4, 5);
addEdge(2, 5, 7);
// Initialising lca and dist values
// with -1 and 0 respectively
for(let i = 1; i <= n; i++)
{
lca[i] = new Array(log);
dist[i] = new Array(log);
for(let j = 0; j < log; j++)
{
lca[i][j] = -1;
dist[i][j] = 0;
}
}
// Perform DFS
dfs(1, -1, 0, 0);
// Query 1: {1, 3}
findDistance(1, 3);
// Query 2: {2, 3}
findDistance(2, 3);
// Query 3: {3, 5}
findDistance(3, 5);
// This code is contributed by decode2207.
</script>
Output:
3 5 12
Time Complexity: The time taken in pre-processing is O(N logN) and every query takes O(logN) time. Therefore, overall time complexity of the solution is O(N logN).
Space Complexity: O(N*log(N))
We are storing the LCA and distance of all the nodes in two 2-D arrays.