LCA in a tree using Binary Lifting Technique

Given a binary tree, the task is to find the Lowest Common Ancestor of the given two nodes in the tree.
Let G be a tree then LCA of two nodes u and v is defined as the node w in the tree which is an ancestor of both u and v and is farthest from the root node.If one node is the ancestor of another one than that particular node is the LCA of those two nodes.
Example:

Input:

Output:
The LCA of 6 and 9 is 1.
The LCA of 5 and 9 is 1.
The LCA of 6 and 8 is 3.
The LCA of 6 and 1 is 1.

Approach: The article describes an approach known as Binary Lifting to find the Lowest Common Ancestor of two nodes in a tree. There can be many approaches to solve the LCA problem. We are discussing the Binary Lifting Technique, the others can be read from here and here.
Binary Lifting is a dynamic programming approach where we pre-compute an array memo[1, n][1, log(n)] where memo[i][j] contains 2^j-th ancestor of node i. For computing the values of memo[][], the following recursion may be used

memo[i][j] = parent[i] if j = 0 and
memo[i][j] = memo[memo[i][j – 1]][j – 1] if j > 0.

We first check whether a node is an ancestor of other or not and if one node is ancestor of other then it is the LCA of these two nodes otherwise we find a node which is not the common ancestor of both u and v and is highest(i.e. a node x such that x is not the common ancestor of u and v but memo[x][0] is) in the tree. After finding such a node (let it be x), we print the first ancestor of x i.e. memo[x][0] which will be the required LCA.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach  
#include <bits/stdc++.h>  
using namespace std;  
  
// Pre-processing to calculate values of memo[][]  
void dfs(int u, int p, int **memo, int *lev, int log, vector<int> *g)  
{  
    // Using recursion formula to calculate  
    // the values of memo[][]  
    memo[u][0] = p;  
    for (int i = 1; i <= log; i++)  
        memo[u][i] = memo[memo[u][i - 1]][i - 1];  
    for (int v : g[u])  
    {  
        if (v != p)  
        {  
            lev[v] = lev[u] + 1;  
            dfs(v, u, memo, lev, log, g);  
        }  
    }  
}  
  
// Function to return the LCA of nodes u and v  
int lca(int u, int v, int log, int *lev, int **memo)  
{  
    // The node which is present farthest  
    // from the root node is taken as u  
    // If v is farther from root node  
    // then swap the two  
    if (lev[u] < lev[v])  
        swap(u, v);  
  
    // Finding the ancestor of u  
    // which is at same level as v  
    for (int i = log; i >= 0; i--)  
        if ((lev[u] - pow(2, i)) >= lev[v])  
            u = memo[u][i];  
  
    // If v is the ancestor of u  
    // then v is the LCA of u and v  
    if (u == v)  
        return u;  
  
    // 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 memo[x][0] is  
    for (int i = log; i >= 0; i--)  
    {  
        if (memo[u][i] != memo[v][i])  
        {  
            u = memo[u][i];  
            v = memo[v][i];  
        }  
    }  
  
    // Returning the first ancestor  
    // of above found node  
    return memo[u][0];  
}  
  
// Driver Code  
int main()  
{  
    // Number of nodes  
    int n = 9;  
  
    // vector to store tree  
    vector<int> g[n + 1];  
  
    int log = (int)ceil(log2(n));  
    int **memo = new int *[n + 1];  
    for (int i = 0; i < n + 1; i++)  
        memo[i] = new int[log + 1];  
  
    // Stores the level of each node  
    int *lev = new int[n + 1];  
  
    // Initialising memo values with -1  
    for (int i = 0; i <= n; i++)  
        memset(memo[i], -1, sizeof memo[i]);  
  
    // Add edges  
    g[1].push_back(2);  
    g[2].push_back(1);  
    g[1].push_back(3);  
    g[3].push_back(1);  
    g[1].push_back(4);  
    g[4].push_back(1);  
    g[2].push_back(5);  
    g[5].push_back(2);  
    g[3].push_back(6);  
    g[6].push_back(3);  
    g[3].push_back(7);  
    g[7].push_back(3);  
    g[3].push_back(8);  
    g[8].push_back(3);  
    g[4].push_back(9);  
    g[9].push_back(4);  
    dfs(1, 1, memo, lev, log, g);  
    cout << "The LCA of 6 and 9 is " << lca(6, 9, log, lev, memo) << endl;  
    cout << "The LCA of 5 and 9 is " << lca(5, 9, log, lev, memo) << endl;  
    cout << "The LCA of 6 and 8 is " << lca(6, 8, log, lev, memo) << endl;  
    cout << "The LCA of 6 and 1 is " << lca(6, 1, log, lev, memo) << endl;  
  
    return 0;  
}  
  
// This code is contributed by  
// sanjeev2552  

Java

// Java implementation of the approach  
import java.util.*;  
public class GFG {  
  
    // ArrayList to store tree  
    static ArrayList<Integer> g[];  
    static int memo[][], lev[], log;  
  
    // Pre-processing to calculate values of memo[][]  
    static void dfs(int u, int p)  
    {  
  
        // Using recursion formula to calculate  
        // the values of memo[][]  
        memo[u][0] = p;  
        for (int i = 1; i <= log; i++)  
            memo[u][i] = memo[memo[u][i - 1]][i - 1];  
        for (int v : g[u]) {  
            if (v != p) {  
  
                // Calculating the level of each node  
                lev[v] = lev[u] + 1;  
                dfs(v, u);  
            }  
        }  
    }  
  
    // Function to return the LCA of nodes u and v  
    static int lca(int u, int v)  
    {  
        // The node which is present farthest  
        // from the root node is taken as u  
        // If v is farther from root node  
        // then swap the two  
        if (lev[u] < lev[v]) {  
            int temp = u;  
            u = v;  
            v = temp;  
        }  
  
        // Finding the ancestor of u  
        // which is at same level as v  
        for (int i = log; i >= 0; i--) {  
            if ((lev[u] - (int)Math.pow(2, i)) >= lev[v])  
                u = memo[u][i];  
        }  
  
        // If v is the ancestor of u  
        // then v is the LCA of u and v  
        if (u == v)  
            return u;  
  
        // 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 memo[x][0] is  
        for (int i = log; i >= 0; i--) {  
            if (memo[u][i] != memo[v][i]) {  
                u = memo[u][i];  
                v = memo[v][i];  
            }  
        }  
  
        // Returning the first ancestor  
        // of above found node  
        return memo[u][0];  
    }  
  
    // Driver code  
    public static void main(String args[])  
    {  
  
        // Number of nodes  
        int n = 9;  
        g = new ArrayList[n + 1];  
  
        // log(n) with base 2  
        log = (int)Math.ceil(Math.log(n) / Math.log(2));  
        memo = new int[n + 1][log + 1];  
  
        // Stores the level of each node  
        lev = new int[n + 1];  
  
        // Initialising memo values with -1  
        for (int i = 0; i <= n; i++)  
            Arrays.fill(memo[i], -1);  
        for (int i = 0; i <= n; i++)  
            g[i] = new ArrayList<>();  
  
        // Add edges  
        g[1].add(2);  
        g[2].add(1);  
        g[1].add(3);  
        g[3].add(1);  
        g[1].add(4);  
        g[4].add(1);  
        g[2].add(5);  
        g[5].add(2);  
        g[3].add(6);  
        g[6].add(3);  
        g[3].add(7);  
        g[7].add(3);  
        g[3].add(8);  
        g[8].add(3);  
        g[4].add(9);  
        g[9].add(4);  
        dfs(1, 1);  
        System.out.println("The LCA of 6 and 9 is " + lca(6, 9));  
        System.out.println("The LCA of 5 and 9 is " + lca(5, 9));  
        System.out.println("The LCA of 6 and 8 is " + lca(6, 8));  
        System.out.println("The LCA of 6 and 1 is " + lca(6, 1));  
    }  
}  

Python3

# Python3 implementation of the above approach  
import math 
  
# Pre-processing to calculate values of memo[][] 
def dfs(u, p, memo, lev, log, g): 
      
    # Using recursion formula to calculate  
    # the values of memo[][]  
    memo[u][0] = p 
    for i in range(1, log + 1): 
        memo[u][i] = memo[memo[u][i - 1]][i - 1] 
          
    for v in g[u]: 
        if v != p: 
            lev[v] = lev[u] + 1
            dfs(v, u, memo, lev, log, g) 
  
# Function to return the LCA of nodes u and v  
def lca(u, v, log, lev, memo): 
      
    # The node which is present farthest  
    # from the root node is taken as u  
    # If v is farther from root node  
    # then swap the two  
    if lev[u] < lev[v]: 
        swap(u, v) 
          
    # Finding the ancestor of u  
    # which is at same level as v  
    for i in range(log, -1, -1): 
        if (lev[u] - pow(2, i)) >= lev[v]: 
            u = memo[u][i] 
              
    # If v is the ancestor of u  
    # then v is the LCA of u and v          
    if u == v: 
        return v 
          
    # 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 memo[x][0] is  
    for i in range(log, -1, -1): 
        if memo[u][i] != memo[v][i]: 
            u = memo[u][i] 
            v = memo[v][i] 
      
    # Returning the first ancestor  
    # of above found node          
    return memo[u][0] 
  
# Driver code  
  
# Number of nodes  
n = 9
  
log = math.ceil(math.log(n, 2)) 
g = [[] for i in range(n + 1)] 
  
memo = [[-1 for i in range(log + 1)]  
            for j in range(n + 1)] 
  
# Stores the level of each node              
lev = [0 for i in range(n + 1)] 
  
# Add edges  
g[1].append(2) 
g[2].append(1) 
g[1].append(3) 
g[3].append(1) 
g[1].append(4) 
g[4].append(1) 
g[2].append(5) 
g[5].append(2) 
g[3].append(6) 
g[6].append(3) 
g[3].append(7) 
g[7].append(3) 
g[3].append(8) 
g[8].append(3) 
g[4].append(9) 
g[9].append(4) 
  
dfs(1, 1, memo, lev, log, g) 
  
print("The LCA of 6 and 9 is", lca(6, 9, log, lev, memo)) 
print("The LCA of 5 and 9 is", lca(5, 9, log, lev, memo)) 
print("The LCA of 6 and 8 is", lca(6, 8, log, lev, memo)) 
print("The LCA of 6 and 1 is", lca(6, 1, log, lev, memo)) 
  
# This code is contributed by Bhaskar 

C#

// C# implementation of the approach  
using System;  
using System.Collections.Generic;  
  
class GFG  
{  
  
    // List to store tree  
    static List<int> []g;  
    static int [,]memo;  
    static int []lev;  
    static int log;  
  
    // Pre-processing to calculate  
    // values of memo[,]  
    static void dfs(int u, int p)  
    {  
  
        // Using recursion formula to  
        // calculate the values of memo[,]  
        memo[u, 0] = p;  
        for (int i = 1; i <= log; i++)  
            memo[u, i] = memo[memo[u, i - 1],  
                                    i - 1];  
        foreach (int v in g[u])  
        {  
            if (v != p)  
            {  
  
                // Calculating the level of each node  
                lev[v] = lev[u] + 1;  
                dfs(v, u);  
            }  
        }  
    }  
  
    // Function to return the LCA of  
    // nodes u and v  
    static int lca(int u, int v)  
    {  
        // The node which is present farthest  
        // from the root node is taken as u  
        // If v is farther from root node  
        // then swap the two  
        if (lev[u] < lev[v])  
        {  
            int temp = u;  
            u = v;  
            v = temp;  
        }  
  
        // Finding the ancestor of u  
        // which is at same level as v  
        for (int i = log; i >= 0; i--)  
        {  
            if ((lev[u] - (int)Math.Pow(2, i)) >= lev[v])  
                u = memo[u, i];  
        }  
  
        // If v is the ancestor of u  
        // then v is the LCA of u and v  
        if (u == v)  
            return u;  
  
        // 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 memo[x,0] is  
        for (int i = log; i >= 0; i--)  
        {  
            if (memo[u, i] != memo[v, i])  
            {  
                u = memo[u, i];  
                v = memo[v, i];  
            }  
        }  
  
        // Returning the first ancestor  
        // of above found node  
        return memo[u, 0];  
    }  
  
    // Driver code  
    public static void Main(String []args)  
    {  
  
        // Number of nodes  
        int n = 9;  
        g = new List<int>[n + 1];  
  
        // log(n) with base 2  
        log = (int)Math.Ceiling(Math.Log(n) / Math.Log(2));  
        memo = new int[n + 1, log + 1];  
  
        // Stores the level of each node  
        lev = new int[n + 1];  
  
        // Initialising memo values with -1  
        for (int i = 0; i <= n; i++)  
            for (int j = 0; j <= log; j++)  
                memo[i, j] = -1;  
        for (int i = 0; i <= n; i++)  
            g[i] = new List<int>();  
  
        // Add edges  
        g[1].Add(2);  
        g[2].Add(1);  
        g[1].Add(3);  
        g[3].Add(1);  
        g[1].Add(4);  
        g[4].Add(1);  
        g[2].Add(5);  
        g[5].Add(2);  
        g[3].Add(6);  
        g[6].Add(3);  
        g[3].Add(7);  
        g[7].Add(3);  
        g[3].Add(8);  
        g[8].Add(3);  
        g[4].Add(9);  
        g[9].Add(4);  
        dfs(1, 1);  
        Console.WriteLine("The LCA of 6 and 9 is " +  
                                        lca(6, 9));  
        Console.WriteLine("The LCA of 5 and 9 is " +  
                                        lca(5, 9));  
        Console.WriteLine("The LCA of 6 and 8 is " +  
                                        lca(6, 8));  
        Console.WriteLine("The LCA of 6 and 1 is " +  
                                        lca(6, 1));  
    }  
}  
  
// This code is contributed by PrinciRaj1992  

Output:

The LCA of 6 and 9 is 1
The LCA of 5 and 9 is 1
The LCA of 6 and 8 is 3
The LCA of 6 and 1 is 1

Time Complexity: The time taken in pre-processing is O(NlogN) and every query takes O(logN) time. So the overall time complexity of the solution is O(NlogN).

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My Personal Notes

C++

Java

Python3

C#

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