# 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:

• In a BIT, root is always node 0. In first level, all nodes are of power of 2 . (1, 2, 4, 8, ….)
• Consider any node in the first level (1, 2, 4, 8, ) its sub-tree will include all the nodes which has same number of bits as that of the root.
1. Sub-Tree with root 1 will have no child.
2. Sub-Tree with root 2 will have 3 as a child.
3. Sub-Tree with root 4 will have 5, 6, 7 as a child.
4. Sub-Tree with root 8 will have 9, 10, 11, 12, 13, 14, 15 as a child. (Double the size of the previous subtree)
5. So subtree with root K will have K nodes including root. And the height of each subtree would be equal:
• for subtree with root 1
• for subtree with root 2
• for subtree with root 4
• Now, we need to find the subtree in which N lies. Say, the height of subtree just before the subtree in which N lies is H and size is L. So, the following cases are possible :
• Case 1 : When N >= L*2 – 1, in such a scenario N is in last level of its subtree. Thus, the diameter will be 2*H + 1. (Path from the lowest level leaf of the previous subtree to the N ).
• Case 2 : When N >= L + L/2 – 1, in such a scenario N is at level H in its subtree. Thus, diameter will be 2*H.
• Case 3 : Otherwise, it is optimal to consider the maximum path length between leaf nodes of two subtree just before the subtree in which N lies i.e diameter is 2*H – 1.

Below are the implementation of the above approach:

## C++

 `#include ` `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; ` `} `

## Java

 `// 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 `

## Python3

 `# 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 `

## C#

 `// 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 `

Output:

```7
```

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