Shortest distance between two nodes in an infinite binary tree

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Consider you have an infinitely long binary tree having a pattern as below:

  
               1
            /      \
           2        3
         /  \      / \
        4    5    6   7
      /  \  / \  / \ / \
     .  .  .  . .  .  .  .

Given two nodes with values x and y. The task is to find the length of the shortest path between the two nodes.

Examples:

Input:  x = 2, y = 3
Output: 2

Input: x = 4, y = 6
Output: 4

A naive approach is to store all the ancestors of both nodes in 2 Data-structures(vectors, arrays, etc..) and do a binary search for the first element(let index i) in vector1, and check if it exists in the vector2 or not. If it does, return the index(let x) of the element in vector2.

The answer will be thus

distance = v1.size() – 1 – i + v2.size() – 1 – x

Below is the implementation of the above approach.

C++

// C++ program to find distance
// between two nodes
// in a infinite binary tree
#include <bits/stdc++.h>
using namespace std;
 
// to stores ancestors of first given node
vector<int> v1;
// to stores ancestors of first given node
vector<int> v2;
 
// normal binary search to find the element
int BinarySearch(int x)
{
    int low = 0;
    int high = v2.size() - 1;
 
    while (low <= high) {
        int mid = (low + high) / 2;
 
        if (v2[mid] == x)
            return mid;
        else if (v2[mid] > x)
            high = mid - 1;
        else
            low = mid + 1;
    }
    return -1;
}
 
// function to make ancestors of first node
void MakeAncestorNode1(int x)
{
    v1.clear();
    while (x) {
        v1.push_back(x);
        x /= 2;
    }
    reverse(v1.begin(), v1.end());
}
 
// function to make ancestors of second node
void MakeAncestorNode2(int x)
{
    v2.clear();
    while (x) {
        v2.push_back(x);
        x /= 2;
    }
    reverse(v2.begin(), v2.end());
}
 
// function to find distance between two nodes
int Distance()
{
    for (int i = v1.size() - 1; i >= 0; i--) {
        int x = BinarySearch(v1[i]);
        if (x != -1) {
            return v1.size() - 1 - i + v2.size() - 1 - x;
        }
    }
}
 
// Driver code
int main()
{
    int node1 = 2, node2 = 3;
 
    // find ancestors
    MakeAncestorNode1(node1);
    MakeAncestorNode2(node2);
 
    cout << "Distance between " << node1 <<
    " and " << node2 << " is : " << Distance();
 
    return 0;
}

Java

// Java program to find distance
// between two nodes
// in a infinite binary tree
import java.util.*;
class GFG 
{
 
// to stores ancestors of first given node
static Vector<Integer> v1 = new Vector<Integer>();
 
// to stores ancestors of first given node
static Vector<Integer> v2 = new Vector<Integer>();
 
// normal binary search to find the element
static int BinarySearch(int x)
{
    int low = 0;
    int high = v2.size() - 1;
 
    while (low <= high)
    {
        int mid = (low + high) / 2;
 
        if (v2.get(mid) == x)
            return mid;
        else if (v2.get(mid) > x)
            high = mid - 1;
        else
            low = mid + 1;
    }
    return -1;
}
 
// function to make ancestors of first node
static void MakeAncestorNode1(int x)
{
    v1.clear();
    while (x > 0)
    {
        v1.add(x);
        x /= 2;
    }
    Collections.reverse(v1);
}
 
// function to make ancestors of second node
static void MakeAncestorNode2(int x)
{
    v2.clear();
    while (x > 0) 
    {
        v2.add(x);
        x /= 2;
    }
    Collections.reverse(v2);
}
 
// function to find distance between two nodes
static int Distance()
{
    for (int i = v1.size() - 1; i >= 0; i--)
    {
        int x = BinarySearch(v1.get(i));
        if (x != -1)
        {
            return v1.size() - 1 - i +
                   v2.size() - 1 - x;
        }
    }
    return Integer.MAX_VALUE;
}
 
// Driver code
public static void main(String[] args) 
{
    int node1 = 2, node2 = 3;
 
    // find ancestors
    MakeAncestorNode1(node1);
    MakeAncestorNode2(node2);
 
    System.out.print("Distance between " + node1 +
                      " and " + node2 + " is : " + 
                                      Distance());
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the distance between 
# two nodes in an infinite binary tree 
 
# normal binary search to find the element 
def BinarySearch(x): 
 
    low = 0
    high = len(v2) - 1
 
    while low <= high: 
        mid = (low + high) // 2
 
        if v2[mid] == x:
            return mid 
        elif v2[mid] > x: 
            high = mid - 1
        else:
            low = mid + 1
     
    return -1
 
# Function to make ancestors of first node 
def MakeAncestorNode1(x): 
 
    v1.clear() 
    while x:
        v1.append(x) 
        x //= 2
     
    v1.reverse() 
 
# Function to make ancestors of second node 
def MakeAncestorNode2(x):
 
    v2.clear() 
    while x:
        v2.append(x) 
        x //= 2
     
    v2.reverse() 
 
# Function to find distance between two nodes 
def Distance(): 
 
    for i in range(len(v1) - 1, -1, -1): 
        x = BinarySearch(v1[i]) 
         
        if x != -1: 
            return (len(v1) - 1 - i +
                    len(v2) - 1 - x) 
     
# Driver code 
if __name__ == "__main__": 
 
    node1, node2 = 2, 3
    v1, v2 = [], []
     
    # Find ancestors 
    MakeAncestorNode1(node1) 
    MakeAncestorNode2(node2) 
 
    print("Distance between", node1, 
          "and", node2, "is :", Distance()) 
 
# This code is contributed by Rituraj Jain

C#

// C# program to find distance
// between two nodes
// in a infinite binary tree
using System;
using System.Collections.Generic;
 
class GFG 
{
 
// to stores ancestors of first given node
static List<int> v1 = new List<int>();
 
// to stores ancestors of first given node
static List<int> v2 = new List<int>();
 
// normal binary search to find the element
static int BinarySearch(int x)
{
    int low = 0;
    int high = v2.Count - 1;
 
    while (low <= high)
    {
        int mid = (low + high) / 2;
 
        if (v2[mid] == x)
            return mid;
        else if (v2[mid] > x)
            high = mid - 1;
        else
            low = mid + 1;
    }
    return -1;
}
 
// function to make ancestors of first node
static void MakeAncestorNode1(int x)
{
    v1.Clear();
    while (x > 0)
    {
        v1.Add(x);
        x /= 2;
    }
    v1.Reverse();
}
 
// function to make ancestors of second node
static void MakeAncestorNode2(int x)
{
    v2.Clear();
    while (x > 0) 
    {
        v2.Add(x);
        x /= 2;
    }
    v2.Reverse();
}
 
// function to find distance between two nodes
static int Distance()
{
    for (int i = v1.Count - 1; i >= 0; i--)
    {
        int x = BinarySearch(v1[i]);
        if (x != -1)
        {
            return v1.Count - 1 - i +
                v2.Count - 1 - x;
        }
    }
    return int.MaxValue;
}
 
// Driver code
public static void Main(String[] args) 
{
    int node1 = 2, node2 = 3;
 
    // find ancestors
    MakeAncestorNode1(node1);
    MakeAncestorNode2(node2);
 
    Console.Write("Distance between " + node1 +
                   " and " + node2 + " is : " + 
                                   Distance());
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
// Javascript program to find distance
// between two nodes
// in a infinite binary tree
 
// to stores ancestors of first given node
let v1 = [];
 
// to stores ancestors of first given node
let v2 = [];
 
// normal binary search to find the element
function BinarySearch(x)
{
    let low = 0;
       let high = v2.length - 1;
  
    while (low <= high)
    {
        let mid = Math.floor((low + high) / 2);
  
        if (v2[mid] == x)
            return mid;
        else if (v2[mid] > x)
            high = mid - 1;
        else
            low = mid + 1;
    }
    return -1;
}
 
// function to make ancestors of first node
function MakeAncestorNode1(x)
{
    v1=[];
    while (x > 0)
    {
        v1.push(x);
        x = Math.floor(x/2);
    }
    v1.reverse();
}
 
// function to make ancestors of second node
function MakeAncestorNode2(x)
{
    v2=[];
    while (x > 0)
    {
        v2.push(x);
        x = Math.floor(x/2);
    }
    v2.reverse();
}
 
// function to find distance between two nodes
function Distance()
{
    for (let i = v1.length - 1; i >= 0; i--)
    {
        let x = BinarySearch(v1[i]);
        if (x != -1)
        {
            return v1.length - 1 - i +
                   v2.length - 1 - x;
        }
    }
    return Number.MAX_VALUE;
}
 
// Driver code
let node1 = 2, node2 = 3;
 
// find ancestors
MakeAncestorNode1(node1);
MakeAncestorNode2(node2);
 
document.write("Distance between " + node1 +
                 " and " + node2 + " is : " +
                 Distance());
 
// This code is contributed by patel2127
</script>

Output

Distance between 2 and 3 is : 2

Complexity Analysis:

Time Complexity: O(log(max(x, y)) * log(max(x, y)))
Auxiliary Space: O(log(max(x, y)))

An efficient approach is to use the property of 2*x and 2*x+1 given. Keep dividing the larger of the two nodes by 2. If the larger becomes the smaller one, then divide the other one. At a stage, both the values will be the same, keep a count on the number of divisions done which will be the answer.

Below is the implementation of the above approach.

C++

// C++ program to find the distance
// between two nodes in an infinite
// binary tree
#include <bits/stdc++.h>
using namespace std;
 
// function to find the distance
// between two nodes in an infinite
// binary tree
int Distance(int x, int y)
{
    // swap the smaller
    if (x < y) {
        swap(x, y);
    }
    int c = 0;
 
    // divide till x!=y
    while (x != y) {
 
        // keep a count
        ++c;
 
        // perform division
        if (x > y)
            x = x >> 1;
 
        // when the smaller
        // becomes the greater
        if (y > x) {
            y = y >> 1;
            ++c;
        }
    }
    return c;
}
 
// Driver code
int main()
{
    int x = 4, y = 6;
    cout << Distance(x, y);
 
 return 0;
}

Java

// Java program to find the distance
// between two nodes in an infinite
// binary tree
class GFG
{
 
// function to find the distance
// between two nodes in an infinite
// binary tree
static int Distance(int x, int y)
{
    // swap the smaller
    if (x < y)
    {
        int temp = x;
        x = y;
        y = temp;
    }
    int c = 0;
 
    // divide till x!=y
    while (x != y) 
    {
 
        // keep a count
        ++c;
 
        // perform division
        if (x > y)
            x = x >> 1;
 
        // when the smaller
        // becomes the greater
        if (y > x) 
        {
            y = y >> 1;
            ++c;
        }
    }
    return c;
}
 
// Driver code
public static void main(String[] args)
{
    int x = 4, y = 6;
    System.out.println(Distance(x, y));
}
}
 
// This code is contributed by PrinciRaj1992 

Python3

# Python3 program to find the distance between
# two nodes in an infinite binary tree 
 
# Function to find the distance between
# two nodes in an infinite binary tree 
def Distance(x, y):
 
    # Swap the smaller 
    if x < y:
        x, y = y, x 
     
    c = 0
     
    # divide till x != y 
    while x != y: 
 
        # keep a count 
        c += 1
 
        # perform division 
        if x > y: 
            x = x >> 1
 
        # when the smaller becomes
        # the greater 
        if y > x: 
            y = y >> 1
            c += 1
     
    return c 
 
# Driver code 
if __name__ == "__main__":
 
    x, y = 4, 6
    print(Distance(x, y)) 
 
# This code is contributed by
# Rituraj Jain

C#

// C# program to find the distance
// between two nodes in an infinite
// binary tree
using System;
 
class GFG
{
 
// function to find the distance
// between two nodes in an infinite
// binary tree
static int Distance(int x, int y)
{
    // swap the smaller
    if (x < y)
    {
        int temp = x;
        x = y;
        y = temp;
    }
    int c = 0;
 
    // divide till x!=y
    while (x != y) 
    {
 
        // keep a count
        ++c;
 
        // perform division
        if (x > y)
            x = x >> 1;
 
        // when the smaller
        // becomes the greater
        if (y > x) 
        {
            y = y >> 1;
            ++c;
        }
    }
    return c;
}
 
// Driver code
public static void Main(String[] args)
{
    int x = 4, y = 6;
    Console.WriteLine(Distance(x, y));
}
}
 
// This code contributed by Rajput-Ji

Javascript

<script>
 
// Javascript program to find the distance
// between two nodes in an infinite
// binary tree
 
// Function to find the distance
// between two nodes in an infinite
// binary tree
function Distance(x, y)
{
     
    // Swap the smaller
    if (x < y)
    {
        let temp = x;
        x = y;
        y = temp;
    }
    let c = 0;
 
    // Divide till x!=y
    while (x != y)
    {
         
        // Keep a count
        ++c;
 
        // Perform division
        if (x > y)
            x = x >> 1;
 
        // When the smaller
        // becomes the greater
        if (y > x)
        {
            y = y >> 1;
            ++c;
        }
    }
    return c;
}
 
// Driver code
let x = 4, y = 6;
 
document.write(Distance(x, y));
 
// This code is contributed by suresh07
 
</script>

Output

Complexity Analysis:

Time Complexity: O(log(max(x, y)))
Auxiliary Space: O(1)

The efficient approach has been suggested by Striver.

Another Approach:

The main idea is to use the formula Level(n) + Level(m) – 2* LCA(n,m) . So Level can easily be calculated using Log base 2 and LCA can be calculated by dividing the greater No. by 2 until n and m become equal.

Below is the implementation of the above approach:

C++

// C++ program to find the distance
// between two nodes in an infinite
// binary tree
#include <bits/stdc++.h>
using namespace std;
 
int LCA(int n,int m)
{
    // swap to keep n smallest
 
    if (n > m) {
        swap(n, m);
    }
 
    // a,b is level of n and m
    int a = log2(n);
    int b = log2(m);
 
    // divide until n!=m
    while (n != m) 
    {
        if (n < m)
            m = m >> 1;
 
        if (n > m)
            n = n >> 1;
    }
 
    // now n==m which is the LCA of n ,m
 
    int v = log2(n);
 
    return  a + b - 2 * v;
}
 
// Driver Code
int main()
{
    int n = 2, m = 6;
     
     // Function call
    cout << LCA(n,m) << endl;
 
    return 0;
}

Java

// Java program to find the distance
// between two nodes in an infinite
// binary tree
import java.util.*;
class GFG{
 
static int LCA(int n,int m)
{
    // swap to keep n smallest
 
    if (n > m) {
        int temp = n;
        n = m;
        m = temp;
    }
 
    // a,b is level of n and m
    int a = (int)(Math.log(n) / Math.log(2));
    int b = (int)(Math.log(m) / Math.log(2));
    // divide until n!=m
    while (n != m) 
    {
        if (n < m)
            m = m >> 1;
 
        if (n > m)
            n = n >> 1;
    }
 
    // now n==m which is the LCA of n ,m
 
    int v = (int)(Math.log(n) / Math.log(2));
    return  a + b - 2 * v;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 2, m = 6;
     
     // Function call
    System.out.print(LCA(n,m) +"\n");
}
}
 
// This code is contributed by umadevi9616 

Python3

# python program to find the distance
# between two nodes in an infinite
# binary tree
import math
def LCA(n, m):
 
    # swap to keep n smallest
    if (n > m):
        n, m = m, n
     
    # a,b is level of n and m
    a = int(math.log2(n))
    b = int(math.log2(m))
 
    # divide until n!=m
    while (n != m):
        if (n < m):
            m = m >> 1
        if (n > m):
            n = n >> 1
     
    # now n==m which is the LCA of n ,m
    v = int(math.log2(n))
    return a + b - 2 * v
 
n = 2
m = 6
     
# Function call
print(LCA(n,m))
 
# This code is contributed by shivanisinghss2110

C#

// C# program to find the distance
// between two nodes in an infinite
// binary tree
using System;
class GFG{
 
static int LCA(int n,int m)
{
    // swap to keep n smallest
    if (n > m) {
        int temp = n;
        n = m;
        m = temp;
    }
 
    // a,b is level of n and m
    int a = (int)(Math.Log(n) / Math.Log(2));
    int b = (int)(Math.Log(m) / Math.Log(2));
   
    // divide until n!=m
    while (n != m) 
    {
        if (n < m)
            m = m >> 1;
 
        if (n > m)
            n = n >> 1;
    }
 
    // now n==m which is the LCA of n ,m
 
    int v = (int)(Math.Log(n) / Math.Log(2));
    return  a + b - 2 * v;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 2, m = 6;
     
     // Function call
    Console.Write(LCA(n,m) +"\n");
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

<script>
 
// JavaScript program to find the distance
// between two nodes in an infinite
// binary tree
function LCA(n, m)
{
     
    // Swap to keep n smallest
    if (n > m)
    {
        let temp = n;
        n = m;
        m = temp;
    }
     
    // a,b is level of n and m
    let a = Math.log2(n);
    let b = Math.log2(m);
     
    // Divide until n!=m
    while (n != m) 
    {
        if (n < m)
            m = m >> 1;
 
        if (n > m)
            n = n >> 1;
    }
 
    // Now n==m which is the LCA of n ,m
    let v = Math.log2(n);
     
    return a + b - 2 * v;
}
 
// Driver Code
let n = 2, m = 6;
 
// Function call
document.write(LCA(n, m));
 
// This code is contributed by shivanisinghss2110
 
</script>

Output

Complexity Analysis:

Time Complexity: O(log(max(x, y)))
Auxiliary Space: O(1)

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