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In this problem, we are given input as the reference to the root of a binary tree. We need to find the diameter of the tree. We solve this using two approaches:
 Approach 1: Using recursion
 Approach 2: Using DFS
A tree is an extensively used data structure in the world of programming. Every node in a tree can have more further subdivisions. The bottommost nodes of a tree that have no subdivisions, are called the leaf nodes.
DIAMETER OF A TREE: It is defined as the number of nodes on the longest path between 2 nodes.This path may or may not pass through the root of the tree.This path includes two leaf nodes. It is also known as the width of a tree.
There can be two possibilities for the longest path of a tree
 It passes through the root.
 It doesn't pass through the root.
In the first example, the diameter is 6, and for the second one, diameter is 5.
There can be more than one longest path, but the diameter will always be maximum.
This problem can be solved with various methods:
Approach 1: Use recursion
We use recursion to calculate the height of a subtree and the diameter. So, we make a recursive function i.e diameter(struct node* tree)
We can consider that the diameter upto any given node will be the sum of the height of its left and right subtrees and 1.
Hence,
Diameter = Left subtree height + Right subtree height + 1.
 If the node that is passed in the recursive function is null, then return zero.
 Calculate the height of the left subtree.
 Calculate the height of the right subtree.
 Calculate the diameter of the left subtree using recursive call till node becomes null.
 Calculate diameter of the right subtree using recursion till node becomes null.
 If the diameter passes through the root node, then the no. of nodes in the path will be : Left subtree height + Right subtree height + 1
 However, if it doesn't pass through the root node, then the diameter will be max(left subtree diameter, right subtree diameter)
 The final diameter will be the maximum value of step 6 and step 7 , and this value will be returned.
This process is continued in recursion till we encounter NULL nodes.
Time complexity : O(n^2)
Space Complexity: O(log n), if a balanced tree, O(n) otherwise. Space complexity is due to recursion.
Code:
//Java program to find the diameter of a Binary Tree
// Structure of the tree
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// Class to print the Diameter
class BinaryTree {
Node root;
// Method to calculate the diameter and return it to main
int diameter(Node root)
{
// base case if tree is empty
if (root == null)
return 0;
// get the height of left and right sub trees
int lheight = height(root.left);
int rheight = height(root.right);
// get the diameter of left and right subtrees
int ldiameter = diameter(root.left);
int rdiameter = diameter(root.right);
/* Return max of following three
1) Diameter of left subtree
2) Diameter of right subtree
3) Height of left subtree + height of right subtree + 1
*/
return Math.max(lheight + rheight + 1,
Math.max(ldiameter, rdiameter));
}
// A wrapper over diameter(Node root)
int diameter() { return diameter(root); }
// function to calculate height of the tree
static int height(Node node)
{
// base case tree is empty
if (node == null)
return 0;
// If tree is not empty then height = 1 + max of
// left height and right heights
return (1
+ Math.max(height(node.left),
height(node.right)));
}
// Driver Code
public static void main(String args[])
{
// creating a binary tree and entering the nodes
BinaryTree tree = new BinaryTree();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
// Function Call
System.out.println(
"The diameter of given binary tree is : "
+ tree.diameter());
}
}
Output:
The diameter of the given binary tree is 4
Approach 2: Using DFS
We can also use the Depth Fisrt Search algorithm to calculate the diameter. Since the longest path always occurs between 2 leaf nodes, if we calculate the farthest node from a leaf node, then we can find the longest path in the tree.
We perform the following steps:
 Start at the root node and find the farthest node from it using DFS.
 Consider this farthest node to be the start node of the longest path.
 Find the farthest node from the start node using DFS.
 This farthest node will be the end node of the longest path.
Time Complexity: O(n)
Space complexity: O(n)
Code:
// Java program to find diameter of a tree using DFS.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Diametre_tree {
// Used to track farthest node.
static int x;
static int maxCount;
static List<Integer> adj[];
// Sets maxCount has maximum distance from node
static void dfsUtil(int node, int count,
boolean visited[],
List<Integer> adj[])
{
visited[node] = true;
count++;
List<Integer> l = adj[node];
for(Integer i: l)
{
if(!visited[i]){
if (count >= maxCount) {
maxCount = count;
x = i;
}
dfsUtil(i, count, visited, adj);
}
}
}
//Recursuve function for DFS traversal
static void dfs(int node, int n, List<Integer>
adj[])
{
boolean[] visited = new boolean[n + 1];
int count = 0;
// Mark all the vertices as not visited
Arrays.fill(visited, false);
// Increment count by 1 for visited node
dfsUtil(node, count + 1, visited, adj);
}
// Returns diameter of binary tree represented
// as adjacency list.
static int diameter(List<Integer> adj[], int n)
{
maxCount = Integer.MIN_VALUE;
/* DFS from a random node and then see
farthest node X from it*/
dfs(1, n, adj);
/* DFS from X and check the farthest node
from it */
dfs(x, n, adj);
return maxCount;
}
//Driver code
public static void main(String args[])
{
int n = 5;
/* Constructed tree is
1
/ \
2 3
/ \
4 5 */
adj = new List[n + 1];
for(int i = 0; i < n+1 ; i++)
adj[i] = new ArrayList<Integer>();
/*create undirected edges */
adj[1].add(2);
adj[2].add(1);
adj[1].add(3);
adj[3].add(1);
adj[2].add(4);
adj[4].add(2);
adj[2].add(5);
adj[5].add(2);
/* maxCount will have diameter of tree */
System.out.println("Diameter of the given tree is " + diameter(adj, n));
}
}
Output:
Diameter of the given tree is 4
Comparison of different solutions
Approach  Time Complexity  Space complexity 

Recursive  O(n^2)  O(n) 
DFS  O(n)  O(n) 
Use cases of finding diameter of a tree:

The average diameter is measured to find the route of direct interprocessor communication, without going through a center, within a network of any structure.

The DADO parallel computer uses a binary tree interconnection network for the rapid execution of rulebased, AIoriented software.