Diameter of a Binary Tree in O(n) [A new method]

The diameter of a tree is the number of nodes on the longest path between two leaves in the tree. The diagram below shows two trees each with diameter nine, the leaves that form the ends of the longest path are colored (note that there may be more than one path in the tree of the same diameter).

Examples:

Input :     1
          /   \
        2      3
      /  \
    4     5

Output : 4

Input :     1
          /   \
        2      3
      /  \ .    \
    4     5 .    6

Output : 5

We have discussed a solution in below post. 
Diameter of a Binary Tree

In this post a new simple O(n) method is discussed. Diameter of a tree can be calculated by only using the height function, because the diameter of a tree is nothing but maximum value of (left_height + right_height + 1) for each node. So we need to calculate this value (left_height + right_height + 1) for each node and update the result. Time complexity – O(n)  

Implementation:

C++

// Simple C++ program to find diameter
// of a binary tree.
#include <bits/stdc++.h>
using namespace std;

/* Tree node structure used in the program */
struct Node {
    int data;
    Node* left, *right;
};

/* 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;
    int ans = INT_MIN; // This will store the final answer
    height(root, ans);
    return ans;
}

struct Node* newNode(int data)
{
    struct Node* node = new Node;
    node->data = data;
    node->left = node->right = NULL;

    return (node);
}

// 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("Diameter is %d\n", diameter(root));

    return 0;
}

Java

// Simple Java program to find diameter
// of a binary tree.
class GfG {

/* Tree node structure used in the program */
static class Node
{
    int data;
    Node left, right;
}

static class A
{
    int ans = Integer.MIN_VALUE;
}

/* Function to find height of a tree */
static int height(Node root, A a)
{
    if (root == null)
        return 0;

    int left_height = height(root.left, a);

    int right_height = height(root.right, a);

    // update the answer, because diameter of a
    // tree is nothing but maximum value of
    // (left_height + right_height + 1) for each node
    a.ans = Math.max(a.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;

    // This will store the final answer
    A a = new A();
    int height_of_tree = height(root, a);
    return a.ans;
}

static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;

    return (node);
}

// 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.println("Diameter is " + diameter(root));
}
}

// This code is contributed by Prerna Saini.

Python3

# Simple Python3 program to find diameter
# of a binary tree.

class newNode:
    def __init__(self, data):
        self.data = data
        self.left = self.right = None

# Function to find height of a tree
def height(root, ans):
    if (root == None):
        return 0

    left_height = height(root.left, ans)

    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[0] = max(ans[0], 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
    ans = [-999999999999] # This will store
                          # the final answer
    height_of_tree = height(root, ans)
    return ans[0]

# Driver code
if __name__ == '__main__':
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)

    print("Diameter is", diameter(root))

# This code is contributed by PranchalK

C#

// Simple C# program to find diameter
// of a binary tree.
using System;

class GfG
{

/* Tree node structure used in the program */
class Node
{
    public int data;
    public Node left, right;
}

class A
{
    public int ans = int.MinValue;
}

/* Function to find height of a tree */
static int height(Node root, A a)
{
    if (root == null)
        return 0;

    int left_height = height(root.left, a);

    int right_height = height(root.right, a);

    // update the answer, because diameter of a
    // tree is nothing but maximum value of
    // (left_height + right_height + 1) for each node
    a.ans = Math.Max(a.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;

    // This will store the final answer
    A a = new A();
    int height_of_tree = height(root, a);
    return a.ans;
}

static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = null;
    node.right = null;

    return (node);
}

// Driver code
public static void Main()
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);

    Console.WriteLine("Diameter is " + diameter(root));
}
}

/* This code is contributed by Rajput-Ji*/

Javascript

<script>

    // Simple Javascript program to find
    // diameter of a binary tree.

    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

    let ans = Number.MIN_VALUE;

    /* 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;

        // This will store the final answer
        let height_of_tree = height(root);
        return ans;
    }

    function newNode(data)
    {
        let node = new Node(data);
        return (node);
    }

    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 is " + diameter(root));

</script>

Output

Diameter is 4

Time complexity: O(n) ,  where n is number of nodes in binary tree .
Auxiliary Space: O(h) for call stack , where h is height of binary tree .

This article is contributed by Pooja Kamal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.