Diameter of a Binary Indexed Tree with N nodes

Given a Binary Indexed Tree with N nodes except root node 0 (Numbered from 1 to N), Find its diameter.

Binary Indexed Tree is a tree where parent of a node number X = X – (X & (X – 1)) i.e. last bit is unset in X. The diameter of a tree is the longest simple path between any two leaves.

Examples:



Input: N = 12
Output: 6
Explanation: Path from node 7 to node 11.

Input : n = 15
Output : 7

Approach:

Below are the implementation of the above approach:

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#include <bits/stdc++.h>
using namespace std;
  
// Function to find diameter
// of BIT with N + 1 nodes
int diameter(int n)
{
    // L is size of subtree just before subtree
    // in which N lies
    int L, H, templen;
    L = 1;
  
    // H is the height of subtree just before
    // subtree in which N lies
    H = 0;
  
    // Base Cases
    if (n == 1) {
        return 1;
    }
    if (n == 2) {
        return 2;
    }
    if (n == 3) {
        return 3;
    }
  
    // Size of subtree are power of 2
    while (L * 2 <= n) {
        L *= 2;
        H++;
    }
  
    // 3 Cases as explained in Approach
    if (n >= L * 2 - 1)
        return 2 * H + 1;
    else if (n >= L + (L / 2) - 1)
        return 2 * H;
    return 2 * H - 1;
}
  
// Driver Code
int main()
{
    int n = 15;
    cout << diameter(n) << endl;
}
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// Java implementation of the approach
class GFG 
{
  
// Function to find diameter
// of BIT with N + 1 nodes
static int diameter(int n)
{
    // L is size of subtree just before subtree
    // in which N lies
    int L, H, templen;
    L = 1;
   
    // H is the height of subtree just before
    // subtree in which N lies
    H = 0;
   
    // Base Cases
    if (n == 1) {
        return 1;
    }
    if (n == 2) {
        return 2;
    }
    if (n == 3) {
        return 3;
    }
   
    // Size of subtree are power of 2
    while (L * 2 <= n) {
        L *= 2;
        H++;
    }
   
    // 3 Cases as explained in Approach
    if (n >= L * 2 - 1)
        return 2 * H + 1;
    else if (n >= L + (L / 2) - 1)
        return 2 * H;
    return 2 * H - 1;
}
   
// Driver Code
public static void main(String []args) 
{
    int n = 15;
  
    System.out.println(diameter(n));
}
}
  
// This code contributed by PrinciRaj1992
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# Python3 implementation of the approach
  
# Function to find diameter
# of BIT with N + 1 nodes
def diameter(n):
      
    # L is size of subtree just before 
    # subtree in which N lies
    L, H, templen = 0, 0, 0;
    L = 1;
  
    # H is the height of subtree just before
    # subtree in which N lies
    H = 0;
  
    # Base Cases
    if (n == 1):
        return 1;
      
    if (n == 2):
        return 2;
      
    if (n == 3):
        return 3;
  
    # Size of subtree are power of 2
    while (L * 2 <= n):
        L *= 2;
        H += 1;
      
    # 3 Cases as explained in Approach
    if (n >= L * 2 - 1):
        return 2 * H + 1;
    elif (n >= L + (L / 2) - 1):
        return 2 * H;
    return 2 * H - 1;
  
# Driver Code
n = 15;
print(diameter(n));
  
# This code is contributed by Rajput-Ji
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// C# implementation of the approach
using System;
  
class GFG 
{
  
// Function to find diameter
// of BIT with N + 1 nodes
static int diameter(int n)
{
    // L is size of subtree just before subtree
    // in which N lies
    int L, H;
    L = 1;
  
    // H is the height of subtree just before
    // subtree in which N lies
    H = 0;
  
    // Base Cases
    if (n == 1)
    {
        return 1;
    }
    if (n == 2) 
    {
        return 2;
    }
    if (n == 3) 
    {
        return 3;
    }
  
    // Size of subtree are power of 2
    while (L * 2 <= n) 
    {
        L *= 2;
        H++;
    }
  
    // 3 Cases as explained in Approach
    if (n >= L * 2 - 1)
        return 2 * H + 1;
    else if (n >= L + (L / 2) - 1)
        return 2 * H;
    return 2 * H - 1;
}
  
// Driver Code
public static void Main(String []args) 
{
    int n = 15;
  
    Console.WriteLine(diameter(n));
}
}
  
// This code is contributed by 29AjayKumar
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Output:
7



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