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Diameter of an N-ary tree

Last Updated : 27 Mar, 2024
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The diameter of an N-ary tree is the longest path present between any two nodes of the tree. These two nodes must be two leaf nodes. The following examples have the longest path[diameter] shaded.

Example 1:

diameternary


Example 2: 

diameterN2

Prerequisite: Diameter of a binary tree.
 
The path can either start from one of the nodes and go up to one of the LCAs of these nodes and again come down to the deepest node of some other subtree or can exist as a diameter of one of the child of the current node. 

The solution will exist in any one of these: 

  1. Diameter of one of the children of the current node 
  2.  Sum of Height of the highest two subtree + 1 

Implementation:

C++ 
// C++ program to find the height of an N-ary
// tree
#include <bits/stdc++.h>
using namespace std;

// Structure of a node of an n-ary tree
struct Node
{
    char key;
    vector<Node *> child;
};

// Utility function to create a new tree node
Node *newNode(int key)
{
    Node *temp = new Node;
    temp->key = key;
    return temp;
}

// Utility function that will return the depth
// of the tree
int depthOfTree(struct Node *ptr)
{
    // Base case
    if (!ptr)
        return 0;

    int maxdepth = 0;

    // Check for all children and find
    // the maximum depth
    for (vector<Node*>::iterator it = ptr->child.begin();
                           it != ptr->child.end(); it++)

        maxdepth = max(maxdepth , depthOfTree(*it));

    return maxdepth + 1;
}

// Function to calculate the diameter
// of the tree
int diameter(struct Node *ptr)
{
    // Base case
    if (!ptr)
        return 0;

    // Find top two highest children
    int max1 = 0, max2 = 0;
    for (vector<Node*>::iterator it = ptr->child.begin();
                          it != ptr->child.end(); it++)
    {
        int h = depthOfTree(*it);
        if (h > max1)
           max2 = max1, max1 = h;
        else if (h > max2)
           max2 = h;
    }

    // Iterate over each child for diameter
    int maxChildDia = 0;
    for (vector<Node*>::iterator it = ptr->child.begin();
                           it != ptr->child.end(); it++)
        maxChildDia = max(maxChildDia, diameter(*it));

    return max(maxChildDia, max1 + max2 + 1);
}

// Driver program
int main()
{
    /*   Let us create below tree
    *           A
    *         / /  \  \
    *       B  F   D  E
    *      / \     |  /|\
    *     K  J    G  C H I
    *      /\            \
    *    N   M            L
    */

    Node *root = newNode('A');
    (root->child).push_back(newNode('B'));
    (root->child).push_back(newNode('F'));
    (root->child).push_back(newNode('D'));
    (root->child).push_back(newNode('E'));
    (root->child[0]->child).push_back(newNode('K'));
    (root->child[0]->child).push_back(newNode('J'));
    (root->child[2]->child).push_back(newNode('G'));
    (root->child[3]->child).push_back(newNode('C'));
    (root->child[3]->child).push_back(newNode('H'));
    (root->child[3]->child).push_back(newNode('I'));
    (root->child[0]->child[0]->child).push_back(newNode('N'));
    (root->child[0]->child[0]->child).push_back(newNode('M'));
    (root->child[3]->child[2]->child).push_back(newNode('L'));

    cout << diameter(root) << endl;

    return 0;
}
Java 
// Java program to find the height of an N-ary
// tree
import java.util.*;
class GFG
{

// Structure of a node of an n-ary tree
static class Node
{
    char key;
    Vector<Node> child;
};

// Utility function to create a new tree node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = (char) key;
    temp.child = new Vector<Node>();
    return temp;
}

// Utility function that will return the depth
// of the tree
static int depthOfTree(Node ptr)
{
    // Base case
    if (ptr == null)
        return 0;

    int maxdepth = 0;

    // Check for all children and find
    // the maximum depth
    for (Node it : ptr.child)

        maxdepth = Math.max(maxdepth,
                            depthOfTree(it));

    return maxdepth + 1;
}

// Function to calculate the diameter
// of the tree
static int diameter(Node ptr)
{
    // Base case
    if (ptr == null)
        return 0;

    // Find top two highest children
    int max1 = 0, max2 = 0;
    for (Node it : ptr.child)
    {
        int h = depthOfTree(it);
        if (h > max1)
        {
            max2 = max1;
            max1 = h;
        }
        else if (h > max2)
        max2 = h;
    }

    // Iterate over each child for diameter
    int maxChildDia = 0;
    for (Node it : ptr.child)
        maxChildDia = Math.max(maxChildDia, 
                               diameter(it));

    return Math.max(maxChildDia, max1 + max2 + 1);
}

// Driver Code
public static void main(String[] args)
{
    /* Let us create below tree
    *         A
    *         / / \ \
    *     B F D E
    *     / \     | /|\
    *     K J G C H I
    *     /\         \
    * N M         L
    */
    Node root = newNode('A');
    (root.child).add(newNode('B'));
    (root.child).add(newNode('F'));
    (root.child).add(newNode('D'));
    (root.child).add(newNode('E'));
    (root.child.get(0).child).add(newNode('K'));
    (root.child.get(0).child).add(newNode('J'));
    (root.child.get(2).child).add(newNode('G'));
    (root.child.get(3).child).add(newNode('C'));
    (root.child.get(3).child).add(newNode('H'));
    (root.child.get(3).child).add(newNode('I'));
    (root.child.get(0).child.get(0).child).add(newNode('N'));
    (root.child.get(0).child.get(0).child).add(newNode('M'));
    (root.child.get(3).child.get(2).child).add(newNode('L'));

    System.out.print(diameter(root) + "\n");
}
}

// This code is contributed by Rajput-Ji
Python3 
# Python program to find the height of an N-ary
# tree

# Structure of a node of an n-ary tree
class Node:
    def __init__(self, x):
        self.key = x
        self.child = []

# Utility function that will return the depth
# of the tree
def depthOfTree(ptr):
    
    # Base case
    if (not ptr):
        return 0
    maxdepth = 0

    # Check for all children and find
    # the maximum depth
    for it in ptr.child:
        maxdepth = max(maxdepth , depthOfTree(it))
    return maxdepth + 1

# Function to calculate the diameter
# of the tree
def diameter(ptr):
    
    # Base case
    if (not ptr):
        return 0

    # Find top two highest children
    max1, max2 = 0, 0
    for it in ptr.child:
        h = depthOfTree(it)
        if (h > max1):
           max2, max1 = max1, h
        elif (h > max2):
           max2 = h

    # Iterate over each child for diameter
    maxChildDia = 0
    for it in ptr.child:
        maxChildDia = max(maxChildDia, diameter(it))
    return max(maxChildDia, max1 + max2 + 1)

# Driver program
if __name__ == '__main__':
    # /*   Let us create below tree
    # *           A
    # *         / /  \  \
    # *       B  F   D  E
    # *      / \     |  /|\
    # *     K  J    G  C H I
    # *      /\            \
    # *    N   M            L
    # */

    root = Node('A')
    (root.child).append(Node('B'))
    (root.child).append(Node('F'))
    (root.child).append(Node('D'))
    (root.child).append(Node('E'))
    (root.child[0].child).append(Node('K'))
    (root.child[0].child).append(Node('J'))
    (root.child[2].child).append(Node('G'))
    (root.child[3].child).append(Node('C'))
    (root.child[3].child).append(Node('H'))
    (root.child[3].child).append(Node('I'))
    (root.child[0].child[0].child).append(Node('N'))
    (root.child[0].child[0].child).append(Node('M'))
    (root.child[3].child[2].child).append(Node('L'))

    print(diameter(root))

# This code is contributed by mohit kumar 29
C# 
// C# program to find the height of 
// an N-ary tree
using System;
using System.Collections.Generic;

class GFG
{

// Structure of a node of an n-ary tree
class Node
{
    public char key;
    public List<Node> child;
};

// Utility function to create 
// a new tree node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = (char) key;
    temp.child = new List<Node>();
    return temp;
}

// Utility function that will return 
// the depth of the tree
static int depthOfTree(Node ptr)
{
    // Base case
    if (ptr == null)
        return 0;

    int maxdepth = 0;

    // Check for all children and find
    // the maximum depth
    foreach (Node it in ptr.child)

        maxdepth = Math.Max(maxdepth,
                            depthOfTree(it));

    return maxdepth + 1;
}

// Function to calculate the diameter
// of the tree
static int diameter(Node ptr)
{
    // Base case
    if (ptr == null)
        return 0;

    // Find top two highest children
    int max1 = 0, max2 = 0;
    foreach (Node it in ptr.child)
    {
        int h = depthOfTree(it);
        if (h > max1)
        {
            max2 = max1;
            max1 = h;
        }
        else if (h > max2)
        max2 = h;
    }

    // Iterate over each child for diameter
    int maxChildDia = 0;
    foreach (Node it in ptr.child)
        maxChildDia = Math.Max(maxChildDia, 
                               diameter(it));

    return Math.Max(maxChildDia, 
                    max1 + max2 + 1);
}

// Driver Code
public static void Main(String[] args)
{
    /* Let us create below tree
    *         A
    *         / / \ \
    *     B F D E
    *     / \     | /|\
    *     K J G C H I
    *     /\         \
    * N M         L
    */
    Node root = newNode('A');
    (root.child).Add(newNode('B'));
    (root.child).Add(newNode('F'));
    (root.child).Add(newNode('D'));
    (root.child).Add(newNode('E'));
    (root.child[0].child).Add(newNode('K'));
    (root.child[0].child).Add(newNode('J'));
    (root.child[2].child).Add(newNode('G'));
    (root.child[3].child).Add(newNode('C'));
    (root.child[3].child).Add(newNode('H'));
    (root.child[3].child).Add(newNode('I'));
    (root.child[0].child[0].child).Add(newNode('N'));
    (root.child[0].child[0].child).Add(newNode('M'));
    (root.child[3].child[2].child).Add(newNode('L'));

    Console.Write(diameter(root) + "\n");
}
}

// This code is contributed by Rajput-Ji
Javascript 
<script>

// Javascript program to find the
// height of an N-ary tree
    
    // Structure of a node of an n-ary tree
    class Node{
        
        // Utility function to create a new tree node
        constructor(key)
        {
            this.key=key;
            this.child=[];
        }
    }
    
    // Utility function that will
    // return the depth
    // of the tree
    function depthOfTree(ptr)
    {
        // Base case
    if (ptr == null)
        return 0;
 
    let maxdepth = 0;
 
    // Check for all children and find
    // the maximum depth
    for (let it=0;it< ptr.child.length;it++)
 
        maxdepth = Math.max(maxdepth,
                     depthOfTree(ptr.child[it]));
 
    return maxdepth + 1;
    }
     
    // Function to calculate the diameter
// of the tree
    function diameter(ptr)
    {
        // Base case
    if (ptr == null)
        return 0;
 
    // Find top two highest children
    let max1 = 0, max2 = 0;
    for (let it=0;it< ptr.child.length;it++)
    {
        let h = depthOfTree(ptr.child[it]);
        if (h > max1)
        {
            max2 = max1;
            max1 = h;
        }
        else if (h > max2)
        max2 = h;
    }
 
    // Iterate over each child for diameter
    let maxChildDia = 0;
    for (let it=0;it< ptr.child.length;it++)
        maxChildDia = Math.max(maxChildDia,
                          diameter(ptr.child[it]));
 
    return Math.max(maxChildDia, max1 + max2 + 1);
    }
    
    // Driver Code
    
    /* Let us create below tree
    *         A
    *         / / \ \
    *     B F D E
    *     / \     | /|\
    *     K J G C H I
    *     /\         \
    * N M         L
    */
    let root = new Node('A');
    (root.child).push(new Node('B'));
    (root.child).push(new Node('F'));
    (root.child).push(new Node('D'));
    (root.child).push(new Node('E'));
    (root.child[0].child).push(new Node('K'));
    (root.child[0].child).push(new Node('J'));
    (root.child[2].child).push(new Node('G'));
    (root.child[3].child).push(new Node('C'));
    (root.child[3].child).push(new Node('H'));
    (root.child[3].child).push(new Node('I'));
    (root.child[0].child[0].child).push(new Node('N'));
    (root.child[0].child[0].child).push(new Node('M'));
    (root.child[3].child[2].child).push(new Node('L'));
 
    document.write(diameter(root) + "\n");

// This code is contributed by patel2127

</script>

Output
7
  • Time Complexity : O( N )
  • Space Complexity : O( N )

Optimizations to above solution:  We can find diameter without calculating depth of the tree making small changes in the above solution, similar to finding diameter of binary tree.

Implementation:

C++ 
// C++ program to find the height of an N-ary
// tree
#include <bits/stdc++.h>
using namespace std;

// Structure of a node of an n-ary tree
struct Node
{
    char key;
    vector<Node *> child;
};

// Utility function to create a new tree node
Node *newNode(int key)
{
    Node *temp = new Node;
    temp->key = key;
    return temp;
}

int diameter(struct Node *ptr,int &diameter_of_tree)
{
    // Base case
    if (!ptr)
        return 0;

    // Find top two highest children
    int max1 = 0, max2 = 0;
    for (vector<Node*>::iterator it = ptr->child.begin();it != ptr->child.end(); it++)
    {
        int h = diameter(*it,diameter_of_tree);
        if (h > max1)
        max2 = max1, max1 = h;
        else if (h > max2)
        max2 = h;
    }

    // Find whether our node can be part of diameter
    diameter_of_tree = max(max1 + max2 + 1,diameter_of_tree);

    return max(max1,max2) + 1;
}

int main()
{
          /* Let us create below tree
         *            A
         *         / / \ \
         *        B F   D E
         *       / \   / /|\
         *      K   J G C H I
         *     /\         |
         *    N  M        L
         */
    Node *root = newNode('A');
    (root->child).push_back(newNode('B'));
    (root->child).push_back(newNode('F'));
    (root->child).push_back(newNode('D'));
    (root->child).push_back(newNode('E'));
    (root->child[0]->child).push_back(newNode('K'));
    (root->child[0]->child).push_back(newNode('J'));
    (root->child[2]->child).push_back(newNode('G'));
    (root->child[3]->child).push_back(newNode('C'));
    (root->child[3]->child).push_back(newNode('H'));
    (root->child[3]->child).push_back(newNode('I'));
    (root->child[0]->child[0]->child).push_back(newNode('N'));
    (root->child[0]->child[0]->child).push_back(newNode('M'));
    (root->child[3]->child[2]->child).push_back(newNode('L'));
    
    // for storing diameter
    int diameter_of_tree = 0;
    
    diameter(root,diameter_of_tree);
    
    cout << diameter_of_tree << endl;

    return 0;
}
// This code is improved by bhuvan
Java 
// Java program to find the height of an N-ary
// tree
import java.util.*;
class GFG {

    // Structure of a node of an n-ary tree
    static class Node {
        char key;
        Vector<Node> child;
    };

    // Utility function to create a new tree node
    static Node newNode(int key)
    {
        Node temp = new Node();
        temp.key = (char)key;
        temp.child = new Vector<Node>();
        return temp;
    }

    // for storing diameter_of_tree
    public static int diameter_of_tree = 0;

    // Function to calculate the diameter
    // of the tree
    static int diameter(Node ptr)
    {
        // Base case
        if (ptr == null)
            return 0;

        // Find top two highest children
        int max1 = 0, max2 = 0;
        for (Node it : ptr.child) {
            int h = diameter(it);
            if (h > max1) {
                max2 = max1;
                max1 = h;
            }
            else if (h > max2)
                max2 = h;
        }

        diameter_of_tree
            = Math.max(max1 + max2 + 1, diameter_of_tree);
        return (Math.max(max1, max2) + 1);
    }

    // Driver Code
    public static void main(String[] args)
    {
        /* Let us create below tree
         *            A
         *         / / \ \
         *        B F   D E
         *       / \   / /|\
         *      K   J G C H I
         *     /\         |
         *    N  M        L
         */
        Node root = newNode('A');
        (root.child).add(newNode('B'));
        (root.child).add(newNode('F'));
        (root.child).add(newNode('D'));
        (root.child).add(newNode('E'));
        (root.child.get(0).child).add(newNode('K'));
        (root.child.get(0).child).add(newNode('J'));
        (root.child.get(2).child).add(newNode('G'));
        (root.child.get(3).child).add(newNode('C'));
        (root.child.get(3).child).add(newNode('H'));
        (root.child.get(3).child).add(newNode('I'));
        (root.child.get(0).child.get(0).child)
            .add(newNode('N'));
        (root.child.get(0).child.get(0).child)
            .add(newNode('M'));
        (root.child.get(3).child.get(2).child)
            .add(newNode('L'));

        diameter(root);
        System.out.print(diameter_of_tree + "\n");
    }
}

// This code is improved by Bhuvan
Python3 
# Python3 program to find the height of an N-ary
# tree

# Structure of a node of an n-ary tree
# Structure of a node of an n-ary tree
class Node:

    # Utility function to create a tree node
    def __init__(self, key):

        self.key = key;
        self.child = [];
diameter_of_tree = 0;

def diameter(ptr):
    global diameter_of_tree

    # Base case
    # Base case
    if (ptr == None):
        return 0;

    # Find top two highest children
    max1 = 0
    max2 = 0;
    for it in range(len(ptr.child)):

        h = diameter(ptr.child[it]);
        if (h > max1):
            max2 = max1
            max1 = h;
        elif (h > max2):
            max2 = h;

    # Find whether our node can be part of diameter
    diameter_of_tree = max(max1 + max2 + 1, diameter_of_tree);

    return max(max1,max2) + 1;

def main():
    ''' us create below tree
        *         A
        *         / / \ \
        *     B F D E
        *     / \ / /|\
        *     K J G C H I
        *     /\         |
        * N M     L
    '''
    root = Node('A');
    (root.child).append(Node('B'));
    (root.child).append(Node('F'));
    (root.child).append(Node('D'));
    (root.child).append(Node('E'));
    (root.child[0].child).append(Node('K'));
    (root.child[0].child).append(Node('J'));
    (root.child[2].child).append(Node('G'));
    (root.child[3].child).append(Node('C'));
    (root.child[3].child).append(Node('H'));
    (root.child[3].child).append(Node('I'));
    (root.child[0].child[0].child).append(Node('N'));
    (root.child[0].child[0].child).append(Node('M'));
    (root.child[3].child[2].child).append(Node('L'));
    
    diameter(root);
    
    print(diameter_of_tree);

main()

# This code is contributed by phasing17.
C# 
// C# program to find the height of an N-ary
// tree

using System;
using System.Collections.Generic;


// Structure of a node of an n-ary tree
class Node {
    public char key;
    public List<Node> child;
};

class GFG {



    // Utility function to create a new tree node
    static Node newNode(int key)
    {
        Node temp = new Node();
        temp.key = (char)key;
        temp.child = new List<Node>();
        return temp;
    }

    // for storing diameter_of_tree
    public static int diameter_of_tree = 0;

    // Function to calculate the diameter
    // of the tree
    static int diameter(Node ptr)
    {
        // Base case
        if (ptr == null)
            return 0;

        // Find top two highest children
        int max1 = 0, max2 = 0;
        foreach (Node it in ptr.child) {
            int h = diameter(it);
            if (h > max1) {
                max2 = max1;
                max1 = h;
            }
            else if (h > max2)
                max2 = h;
        }

        diameter_of_tree
            = Math.Max(max1 + max2 + 1, diameter_of_tree);
        return (Math.Max(max1, max2) + 1);
    }

    // Driver Code
    public static void Main(string[] args)
    {
        /* Let us create below tree
         *            A
         *         / / \ \
         *        B F   D E
         *       / \   / /|\
         *      K   J G C H I
         *     /\         |
         *    N  M        L
         */
        Node root = newNode('A');
        (root.child).Add(newNode('B'));
        (root.child).Add(newNode('F'));
        (root.child).Add(newNode('D'));
        (root.child).Add(newNode('E'));
        (root.child[0].child).Add(newNode('K'));
        (root.child[0].child).Add(newNode('J'));
        (root.child[2].child).Add(newNode('G'));
        (root.child[3].child).Add(newNode('C'));
        (root.child[3].child).Add(newNode('H'));
        (root.child[3].child).Add(newNode('I'));
        (root.child[0].child[0].child)
            .Add(newNode('N'));
        (root.child[0].child[0].child)
            .Add(newNode('M'));
        (root.child[3].child[2].child)
            .Add(newNode('L'));

        diameter(root);
        Console.Write(diameter_of_tree + "\n");
    }
}

// This code is improved by phasing17
Javascript 
// Javascript program to find the height of an N-ary
// tree

// Structure of a node of an n-ary tree
// Structure of a node of an n-ary tree
    class Node{
        
        // Utility function to create a new tree node
        constructor(key)
        {
            this.key=key;
            this.child=[];
        }
    }
    let diameter_of_tree = 0;

function diameter(ptr)
{
    // Base case
    // Base case
    if (ptr == null)
        return 0;

    // Find top two highest children
    let max1 = 0, max2 = 0;
    for (let it=0;it< ptr.child.length;it++)
    {
        let h = diameter(ptr.child[it]);
        if (h > max1)
        max2 = max1, max1 = h;
        else if (h > max2)
        max2 = h;
    }

    // Find whether our node can be part of diameter
    diameter_of_tree = Math.max(max1 + max2 + 1,diameter_of_tree);

    return Math.max(max1,max2) + 1;
}


          /* Let us create below tree
         *            A
         *         / / \ \
         *        B F   D E
         *       / \   / /|\
         *      K   J G C H I
         *     /\         |
         *    N  M        L
         */
    let root = new Node('A');
    (root.child).push(new Node('B'));
    (root.child).push(new Node('F'));
    (root.child).push(new Node('D'));
    (root.child).push(new Node('E'));
    (root.child[0].child).push(new Node('K'));
    (root.child[0].child).push(new Node('J'));
    (root.child[2].child).push(new Node('G'));
    (root.child[3].child).push(new Node('C'));
    (root.child[3].child).push(new Node('H'));
    (root.child[3].child).push(new Node('I'));
    (root.child[0].child[0].child).push(new Node('N'));
    (root.child[0].child[0].child).push(new Node('M'));
    (root.child[3].child[2].child).push(new Node('L'));
    
    diameter(root,diameter_of_tree);
    
    console.log(diameter_of_tree);

  // This code is contributed by garg28harsh.

Output

7
  • Time Complexity: O(N^2)
  • Auxiliary Space: O(N+H) where N is the number of nodes in tree and H is the height of the tree.

A different optimized solution: Longest path in an undirected tree

Another Approach to get diameter using DFS in one traversal:

The diameter of a tree can be calculated as for every node

  • The current node isn’t part of diameter (i.e Diameter lies on one of the children of the current node).
  • The current node is part of diameter (i.e Diameter passes through the current node).

Node: Adjacency List has been used to store the Tree.

Below is the implementation of the above approach:

C++ 
// C++ implementation to find
// diameter of a tree using
// DFS in ONE TRAVERSAL

#include <bits/stdc++.h>
using namespace std;
#define maxN 10005

// The array to store the
// height of the nodes
int height[maxN];

// Adjacency List to store
// the tree
vector<int> tree[maxN];

// variable to store diameter
// of the tree
int diameter = 0;

// Function to add edge between
// node u to node v
void addEdge(int u, int v)
{
    // add edge from u to v
    tree[u].push_back(v);

    // add edge from v to u
    tree[v].push_back(u);
}

void dfs(int cur, int par)
{
    // Variables to store the height of children
    // of cur node with maximum heights
    int max1 = 0;
    int max2 = 0;

    // going in the adjacency list of the current node
    for (auto u : tree[cur]) {
        
        // if that node equals parent discard it
        if (u == par)
            continue;

        // calling dfs for child node
        dfs(u, cur);

        // calculating height of nodes
        height[cur] = max(height[cur], height[u]);

        // getting the height of children
        // of cur node with maximum height
        if (height[u] >= max1) {
            max2 = max1;
            max1 = height[u];
        }
        else if (height[u] > max2) {
            max2 = height[u];
        }
    }

    height[cur] += 1;

    // Diameter of a tree can be calculated as
    // diameter passing through the node
    // diameter doesn't includes the cur node
    diameter = max(diameter, height[cur]);
    diameter = max(diameter, max1 + max2 + 1);
}

// Driver Code
int main()
{
    // n is the number of nodes in tree
    int n = 7;

    // Adding edges to the tree
    addEdge(1, 2);
    addEdge(1, 3);
    addEdge(1, 4);
    addEdge(2, 5);
    addEdge(4, 6);
    addEdge(4, 7);

    // Calling the dfs function to
    // calculate the diameter of tree
    dfs(1, 0);

    cout << "Diameter of tree is : " << diameter - 1
         << "\n";

    return 0;
}
Java 
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;

class GFG {

  static int maxN = 10005; 

  // The array to store the
  // height of the nodes
  static int[] height = new int[maxN];

  // Adjacency List to store
  // the tree
  static ArrayList<ArrayList<Integer>> tree = new ArrayList<ArrayList<Integer>>();

  // variable to store diameter
  // of the tree
  static int diameter = 0;

  // Function to add edge between
  // node u to node v
  static void addEdge(int u, int v)
  {

    // add edge from u to v
    tree.get(u).add(v);

    // add edge from v to u
    tree.get(v).add(u);
  }

  static void dfs(int cur, int par)
  {

    // Variables to store the height of children
    // of cur node with maximum heights
    int max1 = 0;
    int max2 = 0;

    // going in the adjacency list of the current node
    for (int u : tree.get(cur)) {

      // if that node equals parent discard it
      if (u == par)
        continue;

      // calling dfs for child node
      dfs(u, cur);

      // calculating height of nodes
      height[cur] = Math.max(height[cur], height[u]);

      // getting the height of children
      // of cur node with maximum height
      if (height[u] >= max1) {
        max2 = max1;
        max1 = height[u];
      }
      else if (height[u] > max2) {
        max2 = height[u];
      }
    }

    height[cur] += 1;

    // Diameter of a tree can be calculated as
    // diameter passing through the node
    // diameter doesn't includes the cur node
    diameter = Math.max(diameter, height[cur]);
    diameter = Math.max(diameter, max1 + max2 + 1);
  }

  public static void main (String[] args) 
  {

    for(int i = 0; i < maxN; i++)
    {
      tree.add(new ArrayList<Integer>());        
    }

    // n is the number of nodes in tree
    int n = 7;

    // Adding edges to the tree
    addEdge(1, 2);
    addEdge(1, 3);
    addEdge(1, 4);
    addEdge(2, 5);
    addEdge(4, 6);
    addEdge(4, 7);

    // Calling the dfs function to
    // calculate the diameter of tree
    dfs(1, 0);

    System.out.println("Diameter of tree is : " +(diameter - 1));
  }
}

// This code is contributed by ab2127.
Python3 
# C++ implementation to find
# diameter of a tree using
# DFS in ONE TRAVERSAL
maxN = 10005

# The array to store the
# height of the nodes
height = [0 for i in range(maxN)]

# Adjacency List to store
# the tree
tree = [[] for i in range(maxN)]

# variable to store diameter
# of the tree
diameter = 0

# Function to add edge between
# node u to node v
def addEdge(u, v):
  
    # add edge from u to v
    tree[u].append(v)
    
    # add edge from v to u
    tree[v].append(u)

def dfs(cur, par):
    global diameter
    
    # Variables to store the height of children
    # of cur node with maximum heights
    max1 = 0
    max2 = 0
    
    # going in the adjacency list of the current node
    for u in tree[cur]:
      
        # if that node equals parent discard it
        if(u == par):
            continue
        
        # calling dfs for child node
        dfs(u, cur)
        
        # calculating height of nodes
        height[cur] = max(height[cur], height[u])
        
        # getting the height of children
        # of cur node with maximum height
        if(height[u] >= max1):
            max2 = max1
            max1 = height[u]
        elif(height[u] > max2):
            max2 = height[u]
    height[cur] += 1
    
    # Diameter of a tree can be calculated as
    # diameter passing through the node
    # diameter doesn't includes the cur node
    diameter = max(diameter, height[cur])
    diameter = max(diameter, max1 + max2 + 1)

# Driver Code
# n is the number of nodes in tree
n = 7

# Adding edges to the tree
addEdge(1, 2)
addEdge(1, 3)
addEdge(1, 4)
addEdge(2, 5)
addEdge(4, 6)
addEdge(4, 7)

# Calling the dfs function to
# calculate the diameter of tree
dfs(1, 0)

print("Diameter of tree is :", diameter - 1)

# This code is contributed by avanitrachhadiya2155
C# 
using System;
using System.Collections.Generic;

class GFG {

  static int maxN = 10005; 

  // The array to store the
  // height of the nodes
  static int[] height = new int[maxN];

  // Adjacency List to store
  // the tree
  static List<List<int>> tree = new List<List<int>>();

  // variable to store diameter
  // of the tree
  static int diameter = 0;

  // Function to Add edge between
  // node u to node v
  static void AddEdge(int u, int v)
  {

    // Add edge from u to v
    tree[u].Add(v);

    // Add edge from v to u
    tree[v].Add(u);
  }

  static void dfs(int cur, int par)
  {

    // Variables to store the height of children
    // of cur node with maximum heights
    int max1 = 0;
    int max2 = 0;

    // going in the adjacency list of the current node
    foreach (int u in tree[cur]) {

      // if that node equals parent discard it
      if (u == par)
        continue;

      // calling dfs for child node
      dfs(u, cur);

      // calculating height of nodes
      height[cur] = Math.Max(height[cur], height[u]);

      // getting the height of children
      // of cur node with maximum height
      if (height[u] >= max1) {
        max2 = max1;
        max1 = height[u];
      }
      else if (height[u] > max2) {
        max2 = height[u];
      }
    }

    height[cur] += 1;

    // Diameter of a tree can be calculated as
    // diameter passing through the node
    // diameter doesn't includes the cur node
    diameter = Math.Max(diameter, height[cur]);
    diameter = Math.Max(diameter, max1 + max2 + 1);
  }

  public static void Main (string[] args) 
  {

    for(int i = 0; i < maxN; i++)
    {
      tree.Add(new List<int>());        
    }

    // n is the number of nodes in tree
    int n = 7;

    // Adding edges to the tree
    AddEdge(1, 2);
    AddEdge(1, 3);
    AddEdge(1, 4);
    AddEdge(2, 5);
    AddEdge(4, 6);
    AddEdge(4, 7);

    // Calling the dfs function to
    // calculate the diameter of tree
    dfs(1, 0);

    Console.WriteLine("Diameter of tree is : " +(diameter - 1));
  }
}

// This code is contributed by phasing17.
Javascript 
<script>

// Javascript implementation to find
// diameter of a tree using
// DFS in ONE TRAVERSAL

let maxN = 10005;

// The array to store the
// height of the nodes
let height=new Array(maxN);

// Adjacency List to store
// the tree
let tree=new Array(maxN);
for(let i=0;i<maxN;i++)
{
    height[i]=0;
    tree[i]=[];
}

// variable to store diameter
// of the tree
let diameter = 0;

// Function to add edge between
// node u to node v
function addEdge(u,v)
{
    // add edge from u to v
    tree[u].push(v);
 
    // add edge from v to u
    tree[v].push(u);
}

function dfs(cur,par)
{
    // Variables to store the height of children
    // of cur node with maximum heights
    let max1 = 0;
    let max2 = 0;
 
    // going in the adjacency list 
    // of the current node
    for (let u=0;u<tree[cur].length;u++) {
         
        // if that node equals parent discard it
        if (tree[cur][u] == par)
            continue;
 
        // calling dfs for child node
        dfs(tree[cur][u], cur);
 
        // calculating height of nodes
        height[cur] = Math.max(height[cur], 
        height[tree[cur][u]]);
 
        // getting the height of children
        // of cur node with maximum height
        if (height[tree[cur][u]] >= max1) {
            max2 = max1;
            max1 = height[tree[cur][u]];
        }
        else if (height[tree[cur][u]] > max2) {
            max2 = height[tree[cur][u]];
        }
    }
 
    height[cur] += 1;
 
    // Diameter of a tree can be calculated as
    // diameter passing through the node
    // diameter doesn't includes the cur node
    diameter = Math.max(diameter, height[cur]);
    diameter = Math.max(diameter, max1 + max2 + 1);
}

// Driver Code

// n is the number of nodes in tree
let n = 7;

// Adding edges to the tree
addEdge(1, 2);
addEdge(1, 3);
addEdge(1, 4);
addEdge(2, 5);
addEdge(4, 6);
addEdge(4, 7);

// Calling the dfs function to
// calculate the diameter of tree
dfs(1, 0);

document.write("Diameter of tree is : " +(diameter - 1))

    
// This code is contributed by unknown2108

</script>

Output
Diameter of tree is : 4
  • Time Complexity: O(N), Where N is the number of nodes in given binary tree.
  • Auxiliary Space: O(N)


S

Shubham Gupta
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