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# Maximum length cycle that can be formed by joining two nodes of a binary tree

• Difficulty Level : Medium
• Last Updated : 05 Aug, 2021

Given a binary tree, the task is to find the maximum length of the cycle that can be formed by joining any two nodes of the tree.
Examples:

```Input:
1
/  \
2    3
\     \
5     6

Output: 5
Cycle can be formed by joining node with value 5 and 6.

Input:
1
/  \
3    4
/ \
5   6
/     \
7       8
\     /
11   9

Output: 7```

Approach: The idea is to find the diameter of the given binary tree, since cycle with maximum length will be equal to the diameter of the binary tree.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Tree node structure``struct` `Node {``    ``int` `data;``    ``Node *left, *right;``};` `struct` `Node* newNode(``int` `data)``{``    ``struct` `Node* node = ``new` `Node;``    ``node->data = data;``    ``node->left = node->right = NULL;` `    ``return` `(node);``}` `// Function to find height of a tree``int` `height(Node* root, ``int``& ans)``{``    ``if` `(root == NULL)``        ``return` `0;` `    ``int` `left_height = height(root->left, ans);` `    ``int` `right_height = height(root->right, ans);` `    ``// Update the answer, because diameter of a``    ``// tree is nothing but maximum value of``    ``// (left_height + right_height + 1) for each node``    ``ans = max(ans, 1 + left_height + right_height);` `    ``return` `1 + max(left_height, right_height);``}` `// Computes the diameter of binary tree``// with given root``int` `diameter(Node* root)``{``    ``if` `(root == NULL)``        ``return` `0;` `    ``// Variable to store the final answer``    ``int` `ans = INT_MIN;` `    ``int` `height_of_tree = height(root, ans);``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``struct` `Node* root = newNode(1);``    ``root->left = newNode(2);``    ``root->right = newNode(3);``    ``root->left->left = newNode(4);``    ``root->left->right = newNode(5);` `    ``printf``(``"%d"``, diameter(root));` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Tree node structure``static` `class` `Node``{``    ``int` `data;``    ``Node left, right;``};` `static` `int` `ans;``static` `Node newNode(``int` `data)``{``    ``Node node = ``new` `Node();``    ``node.data = data;``    ``node.left = node.right = ``null``;` `    ``return` `(node);``}` `// Function to find height of a tree``static` `int` `height(Node root)``{``    ``if` `(root == ``null``)``        ``return` `0``;` `    ``int` `left_height = height(root.left);` `    ``int` `right_height = height(root.right);` `    ``// Update the answer, because diameter of a``    ``// tree is nothing but maximum value of``    ``// (left_height + right_height + 1) for each node``    ``ans = Math.max(ans, ``1` `+ left_height + right_height);` `    ``return` `1` `+ Math.max(left_height, right_height);``}` `// Computes the diameter of binary tree``// with given root``static` `int` `diameter(Node root)``{``    ``if` `(root == ``null``)``        ``return` `0``;` `    ``// Variable to store the final answer``    ``ans = Integer.MIN_VALUE;` `    ``int` `height_of_tree = height(root);``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``Node root = newNode(``1``);``    ``root.left = newNode(``2``);``    ``root.right = newNode(``3``);``    ``root.left.left = newNode(``4``);``    ``root.left.right = newNode(``5``);` `    ``System.out.printf(``"%d"``, diameter(root));``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation of the approach` `# Tree node structure``class` `Node:``    ` `    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function to find height of a tree``def` `height(root):`` ` `    ``if` `root ``=``=` `None``:``        ``return` `0``        ` `    ``global` `ans``    ``left_height ``=` `height(root.left)``    ``right_height ``=` `height(root.right)` `    ``# Update the answer, because diameter of a``    ``# tree is nothing but maximum value of``    ``# (left_height + right_height + 1) for each node``    ``ans ``=` `max``(ans, ``1` `+` `left_height ``+` `right_height)` `    ``return` `1` `+` `max``(left_height, right_height)` `# Computes the diameter of``# binary tree with given root``def` `diameter(root):`` ` `    ``if` `root ``=``=` `None``:``        ``return` `0` `    ``height_of_tree ``=` `height(root)``    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `"__main__"``:`` ` `    ``root ``=` `Node(``1``)``    ``root.left ``=` `Node(``2``)``    ``root.right ``=` `Node(``3``)``    ``root.left.left ``=` `Node(``4``)``    ``root.left.right ``=` `Node(``5``)``    ` `    ``ans ``=` `0``    ``print``(diameter(root))``    ` `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the approach``using` `System;``    ` `class` `GFG``{` `// Tree node structure``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;``};` `static` `int` `ans;``static` `Node newNode(``int` `data)``{``    ``Node node = ``new` `Node();``    ``node.data = data;``    ``node.left = node.right = ``null``;` `    ``return` `(node);``}` `// Function to find height of a tree``static` `int` `height(Node root)``{``    ``if` `(root == ``null``)``        ``return` `0;` `    ``int` `left_height = height(root.left);` `    ``int` `right_height = height(root.right);` `    ``// Update the answer, because diameter of a``    ``// tree is nothing but maximum value of``    ``// (left_height + right_height + 1) for each node``    ``ans = Math.Max(ans, 1 + left_height +``                            ``right_height);` `    ``return` `1 + Math.Max(left_height,``                        ``right_height);``}` `// Computes the diameter of binary tree``// with given root``static` `int` `diameter(Node root)``{``    ``if` `(root == ``null``)``        ``return` `0;` `    ``// Variable to store the final answer``    ``ans = ``int``.MinValue;` `    ``int` `height_of_tree = height(root);``    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``Node root = newNode(1);``    ``root.left = newNode(2);``    ``root.right = newNode(3);``    ``root.left.left = newNode(4);``    ``root.left.right = newNode(5);` `    ``Console.WriteLine(``"{0}"``, diameter(root));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`4`

Time Complexity: O(N)
Auxiliary Space: O(N)

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