Skip to content
Related Articles

Related Articles

Improve Article

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 <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;
}

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




<script>
 
    // JavaScript implementation of the approach
     
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    let ans;
    function newNode(data)
    {
        let node = new Node(data);
        return (node);
    }
 
    // Function to find height of a tree
    function height(root)
    {
        if (root == null)
            return 0;
 
        let left_height = height(root.left);
 
        let 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
    function diameter(root)
    {
        if (root == null)
            return 0;
 
        // Variable to store the final answer
        ans = Number.MIN_VALUE;
 
        let height_of_tree = height(root);
        return ans;
    }
     
    let root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
   
    document.write(diameter(root));
     
</script>
Output: 
4

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :