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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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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

Output:

4


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Improved By : rituraj_jain