Diameter of a Binary Tree
Last Updated :
17 Apr, 2024
The diameter/width of a tree is defined as the number of nodes on the longest path between two end nodes.Â
The diagram below shows two trees each with a diameter of nine, the leaves that form the ends of the longest path are shaded (note that there is more than one path in each tree of length nine, but no path longer than nine nodes).Â
Approach: The diameter of a tree T is the largest of the following quantities:
- The diameter of T’s left subtree.
- The diameter of T’s right subtree.
- The longest path between leaves that goes through the root of T (this can be computed from the heights of the subtrees of T)
Below is the implementation of the above approach
C++
// Recursive optimized C program to find the diameter of a
// Binary Tree
#include <bits/stdc++.h>
using namespace std;
// A binary tree node has data, pointer to left child
// and a pointer to right child
struct node {
int data;
struct node *left, *right;
};
// function to create a new node of tree and returns pointer
struct node* newNode(int data);
// returns max of two integers
int max(int a, int b) { return (a > b) ? a : b; }
// function to Compute height of a tree.
int height(struct node* node);
// Function to get diameter of a binary tree
int diameter(struct node* tree)
{
// base case where tree is empty
if (tree == NULL)
return 0;
// get the height of left and right sub-trees
int lheight = height(tree->left);
int rheight = height(tree->right);
// get the diameter of left and right sub-trees
int ldiameter = diameter(tree->left);
int rdiameter = diameter(tree->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 max(lheight + rheight + 1,
max(ldiameter, rdiameter));
}
// UTILITY FUNCTIONS TO TEST diameter() FUNCTION
// The function Compute the "height" of a tree. Height is
// the number f nodes along the longest path from the root
// node down to the farthest leaf node.
int height(struct 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 + max(height(node->left), height(node->right));
}
// Helper function that allocates a new node with the
// given data and NULL left and right pointers.
struct node* newNode(int data)
{
struct node* node
= (struct node*)malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
// Driver Code
int main()
{
/* Constructed binary tree is
1
/ \
2 3
/ \
4 5
*/
struct node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
// Function Call
cout << "Diameter of the given binary tree is "
<< diameter(root);
return 0;
}
// This code is contributed by shivanisinghss2110
// Recursive optimized C program to find the diameter of a
// Binary Tree
#include <stdio.h>
#include <stdlib.h>
// A binary tree node has data, pointer to left child
// and a pointer to right child
struct node {
int data;
struct node *left, *right;
};
// function to create a new node of tree and returns pointer
struct node* newNode(int data);
// returns max of two integers
int max(int a, int b) { return (a > b) ? a : b; }
// function to Compute height of a tree.
int height(struct node* node);
// Function to get diameter of a binary tree
int diameter(struct node* tree)
{
// base case where tree is empty
if (tree == NULL)
return 0;
// get the height of left and right sub-trees
int lheight = height(tree->left);
int rheight = height(tree->right);
// get the diameter of left and right sub-trees
int ldiameter = diameter(tree->left);
int rdiameter = diameter(tree->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 max(lheight + rheight + 1,
max(ldiameter, rdiameter));
}
// UTILITY FUNCTIONS TO TEST diameter() FUNCTION
// The function Compute the "height" of a tree. Height is
// the number f nodes along the longest path from the root
// node down to the farthest leaf node.
int height(struct 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 + max(height(node->left), height(node->right));
}
// Helper function that allocates a new node with the
// given data and NULL left and right pointers.
struct node* newNode(int data)
{
struct node* node
= (struct node*)malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
// Driver Code
int main()
{
/* Constructed binary tree is
1
/ \
2 3
/ \
4 5
*/
struct node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
// Function Call
printf("Diameter of the given binary tree is %d\n",
diameter(root));
return 0;
}
// Recursive optimized Java program to find the diameter of
// a Binary Tree
// Class containing left and right child of current
// node and key value
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 sub-trees
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); }
// The function Compute the "height" of a tree. Height
// is the number of nodes along the longest path from
// the root node down to the farthest leaf node.
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());
}
}
# Python3 program to find the diameter of binary tree
# A binary tree node
class Node:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# The function Compute the "height" of a tree. Height is the
# number of nodes along the longest path from the root node
# down to the farthest leaf node.
def height(node):
# Base Case : Tree is empty
if node is None:
return 0
# If tree is not empty then height = 1 + max of left
# height and right heights
return 1 + max(height(node.left), height(node.right))
# Function to get the diameter of a binary tree
def diameter(root):
# Base Case when tree is empty
if root is None:
return 0
# Get the height of left and right sub-trees
lheight = height(root.left)
rheight = height(root.right)
# Get the diameter of left and right sub-trees
ldiameter = diameter(root.left)
rdiameter = diameter(root.right)
# Return max of the following tree:
# 1) Diameter of left subtree
# 2) Diameter of right subtree
# 3) Height of left subtree + height of right subtree +1
return max(lheight + rheight + 1, max(ldiameter, rdiameter))
# Driver Code
if __name__ == "__main__":
"""
Constructed binary tree is
1
/ \
2 3
/ \
4 5
"""
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
# Function Call
print(diameter(root))
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// Recursive optimized C# program to find the diameter of
// a Binary Tree
// Class containing left and right child of current
// node and key value
using System;
namespace Tree {
class Tree<T> {
public Tree(T value) { this.value = value; }
public T value
{
get;
set;
}
public Tree<T> left
{
get;
set;
}
public Tree<T> right
{
get;
set;
}
}
public class TreeDiameter {
Tree<int> root;
// The function Compute the "height" of a tree. Height
// is the number of nodes along the longest path from
// the root node down to the farthest leaf node.
int Height(Tree<int> node)
{
if (node == null)
return 0;
return 1
+ Math.Max(Height(node.left),
Height(node.right));
}
int Diameter(Tree<int> root)
{
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 sub-trees
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); }
// Driver Code
public static void Main(string[] args)
{
// creating a binary tree and entering the nodes
TreeDiameter tree = new TreeDiameter();
tree.root = new Tree<int>(1);
tree.root.left = new Tree<int>(2);
tree.root.right = new Tree<int>(3);
tree.root.left.left = new Tree<int>(4);
tree.root.left.right = new Tree<int>(5);
Console.WriteLine(
"The diameter of given binary tree is : "
+ tree.Diameter());
}
}
}
// This code is contributed by krishaccot
<script>
// JavaScript program to find the diameter of binary tree
// A binary tree node
class Node{
// Constructor to create a new node
constructor(data){
this.data = data
this.left = null
this.right = null
}
}
// The function Compute the "height" of a tree. Height is the
// number of nodes along the longest path from the root node
// down to the farthest leaf node.
function height(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))
}
// Function to get the diameter of a binary tree
function diameter(root){
// Base Case when tree is empty
if(root == null)
return 0
// Get the height of left and right sub-trees
let lheight = height(root.left)
let rheight = height(root.right)
// Get the diameter of left and right sub-trees
let ldiameter = diameter(root.left)
let rdiameter = diameter(root.right)
// Return max of the following tree:
// 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))
}
// Driver Code
// Constructed binary tree is
// 1
// / \
// 2 3
// / \
// 4 5
root = new Node(1)
root.left = new Node(2)
root.right = new Node(3)
root.left.left = new Node(4)
root.left.right = new Node(5)
// Function Call
document.write("Diameter of the given binary tree is "+diameter(root))
// This code is contributed by shinjanpatra
</script>
OutputDiameter of the given binary tree is 4
Time Complexity: O(N2)
Auxiliary Space: O(N) for call stack
Â
Efficient Approach: To solve the problem follow the below idea:
 The above implementation can be optimized by calculating the height in the same recursion rather than calling a height() separately. Thanks to Amar for suggesting this optimized version.
Below is the implementation of the above approach:
C++
// Recursive optimized C++ program to find the diameter of a
// Binary Tree
#include <bits/stdc++.h>
using namespace std;
// A binary tree node has data, pointer to left child
// and a pointer to right child
struct node {
int data;
struct node *left, *right;
};
// function to create a new node of tree and returns pointer
struct node* newNode(int data);
int diameterOpt(struct node* root, int* height)
{
// lh --> Height of left subtree
// rh --> Height of right subtree
int lh = 0, rh = 0;
// ldiameter --> diameter of left subtree
// rdiameter --> Diameter of right subtree
int ldiameter = 0, rdiameter = 0;
if (root == NULL) {
*height = 0;
return 0; // diameter is also 0
}
// Get the heights of left and right subtrees in lh and
// rh And store the returned values in ldiameter and
// ldiameter
ldiameter = diameterOpt(root->left, &lh);
rdiameter = diameterOpt(root->right, &rh);
// Height of current node is max of heights of left and
// right subtrees plus 1
*height = max(lh, rh) + 1;
return max(lh + rh + 1, max(ldiameter, rdiameter));
}
// Helper function that allocates a new node with the
// given data and NULL left and right pointers.
struct node* newNode(int data)
{
struct node* node
= (struct node*)malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
// Driver Code
int main()
{
/* Constructed binary tree is
1
/ \
2 3
/ \
4 5
*/
struct node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
int height = 0;
// Function Call
cout << "Diameter of the given binary tree is "
<< diameterOpt(root, &height);
return 0;
}
// This code is contributed by probinsah.
// Recursive Java program to find the diameter of a
// Binary Tree
// Class containing left and right child of current
// node and key value
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// A utility class to pass height object
class Height {
int h;
}
// Class to print the Diameter
class BinaryTree {
Node root;
// define height =0 globally and call
// diameterOpt(root,height) from main
int diameterOpt(Node root, Height height)
{
// lh --> Height of left subtree
// rh --> Height of right subtree
Height lh = new Height(), rh = new Height();
if (root == null) {
height.h = 0;
return 0; // diameter is also 0
}
/*
ldiameter --> diameter of left subtree
rdiameter --> Diameter of right subtree
Get the heights of left and right subtrees
in lh and rh. And store the returned values in
ldiameter and ldiameter*/
int ldiameter = diameterOpt(root.left, lh);
int rdiameter = diameterOpt(root.right, rh);
// Height of current node is max of heights of left
// and right subtrees plus 1
height.h = Math.max(lh.h, rh.h) + 1;
return Math.max(lh.h + rh.h + 1,
Math.max(ldiameter, rdiameter));
}
// A wrapper over diameter(Node root)
int diameter()
{
Height height = new Height();
return diameterOpt(root, height);
}
// The function Compute the "height" of a tree. Height
// is
// the number f nodes along the longest path from the
// root node down to the farthest leaf node.
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());
}
}
# Python3 program to find the diameter of a binary tree
# A binary tree Node
class Node:
# Constructor to create a new Node
def __init__(self, data):
self.data = data
self.left = self.right = None
# utility class to pass height object
class Height:
def __init(self):
self.h = 0
# Optimised recursive function to find diameter
# of binary tree
def diameterOpt(root, height):
# to store height of left and right subtree
lh = Height()
rh = Height()
# base condition- when binary tree is empty
if root is None:
height.h = 0
return 0
# ldiameter --> diameter of left subtree
# rdiameter --> diameter of right subtree
# height of left subtree and right subtree is obtained from lh and rh
# and returned value of function is stored in ldiameter and rdiameter
ldiameter = diameterOpt(root.left, lh)
rdiameter = diameterOpt(root.right, rh)
# height of tree will be max of left subtree
# height and right subtree height plus1
height.h = max(lh.h, rh.h) + 1
# return maximum of the following
# 1)left diameter
# 2)right diameter
# 3)left height + right height + 1
return max(lh.h + rh.h + 1, max(ldiameter, rdiameter))
# function to calculate diameter of binary tree
def diameter(root):
height = Height()
return diameterOpt(root, height)
# 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)
"""
Constructed binary tree is
1
/ \
2 3
/ \
4 5
"""
print("The diameter of the binary tree is:", end=" ")
# Function Call
print(diameter(root))
# This code is contributed by Shweta Singh(shweta44)
// Recursive C# program to find the diameter of a
// Binary Tree
using System;
using System.Collections.Generic;
// Class containing left and right child of current
// node and key value
class Node {
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// A utility class to pass height object
class Height {
public int h;
}
// Class to print the Diameter
class BinaryTree {
public Node root;
// define height =0 globally and call
// diameterOpt(root,height) from main
public int diameterOpt(Node root, Height height)
{
// lh --> Height of left subtree
// rh --> Height of right subtree
Height lh = new Height(), rh = new Height();
if (root == null) {
height.h = 0;
return 0; // diameter is also 0
}
// ldiameter --> diameter of left subtree
// rdiameter --> Diameter of right subtree
// Get the heights of left and right subtrees in lh
/*and rh And store the returned values in ldiameter
and ldiameter */
int ldiameter = diameterOpt(root.left, lh);
int rdiameter = diameterOpt(root.right, rh);
// Height of current node is max of heights of left
// and right subtrees plus 1
height.h = Math.Max(lh.h, rh.h) + 1;
return Math.Max(lh.h + rh.h + 1,
Math.Max(ldiameter, rdiameter));
}
// A wrapper over diameter(Node root)
public int diameter()
{
Height height = new Height();
return diameterOpt(root, height);
}
// The function Compute the "height" of a tree. Height
// is
// the number f nodes along the longest path from the
// root node down to the farthest leaf node.
public 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
static void Main()
{
// 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
Console.Write(
"The diameter of given binary tree is : "
+ tree.diameter());
}
}
// This code is contributed by divyesh072019.
<script>
// JavaScript program to find the diameter of a binary tree
// A binary tree Node
class Node{
// Constructor to create a new Node
constructor(data){
this.data = data
this.left = this.right = null
}
}
// utility class to pass height object
class Height
{
constructor()
{
this.h = 0
}
}
// Optimised recursive function to find diameter
// of binary tree
function diameterOpt(root, height){
// to store height of left and right subtree
let lh = new Height()
let rh = new Height()
// base condition- when binary tree is empty
if(root == null)
{
height.h = 0
return 0
}
// ldiameter --> diameter of left subtree
// rdiameter --> diameter of right subtree
// height of left subtree and right subtree is obtained from lh and rh
// and returned value of function is stored in ldiameter and rdiameter
let ldiameter = diameterOpt(root.left, lh)
let rdiameter = diameterOpt(root.right, rh)
// height of tree will be max of left subtree
// height and right subtree height plus1
height.h = Math.max(lh.h, rh.h) + 1
// return maximum of the following
// 1)left diameter
// 2)right diameter
// 3)left height + right height + 1
return Math.max(lh.h + rh.h + 1, Math.max(ldiameter, rdiameter))
}
// function to calculate diameter of binary tree
function diameter(root){
let height = new Height()
return diameterOpt(root, height)
}
// Driver Code
let root = new Node(1)
root.left = new Node(2)
root.right = new Node(3)
root.left.left = new Node(4)
root.left.right = new Node(5)
// Constructed binary tree is
// 1
// / \
// 2 3
// / \
// 4 5
// Function Call
document.write(diameter(root))
// This code is contributed by Shinjanpatra
</script>
OutputDiameter of the given binary tree is 4
Time Complexity: O(N)Â
Auxiliary Space: O(N) due to recursive calls.
The main idea behind Morris Traversal is to modify the binary tree structure by creating a link from the rightmost node of the left subtree to its parent (as shown in the image below), which allows us to traverse the tree without using extra space for a stack or recursion.
.png)
Follow the steps below to implement the above idea:
- Traverse the binary tree using a current_node initialised as the root of the binary tree.
- While the current_node is not NULL, do the following
- If the left child of the current node is NULL, move to the right child.
- If the left child of the current node is not NULL, find the rightmost node in the left subtree of the current node.
- If the right child of the rightmost node is NULL, set its right child to the current node, and move to the left child of the current node.
- If the right child of the rightmost node is not NULL, set its right child back to NULL, visit the current node, and move to its right child.
- For each visited node, calculate its left and right subtree heights using the maximum function and update the diameter as the maximum of the sum of their heights and the current diameter.
- Return the diameter.
Below is the implementation of the above approach:
C++
// C++ code to implement the above approach that uses the
// morris traversal algorithm
#include <algorithm>
#include <iostream>
using namespace std;
// A tree node
struct Node {
int data;
struct Node* left;
struct Node* right;
};
// Create a new node
Node* newNode(int data)
{
Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Morris traversal to find the diameter of the binary tree
int findDiameter(Node* root)
{
if(root == NULL)
return 0;
if(root->left==NULL && root->right==NULL)
return 1;
int ans = 0;
Node* curr = root;
while (curr != NULL) {
if (curr->left == NULL) {
curr = curr->right;
}
else {
Node* pre = curr->left;
while (pre->right != NULL && pre->right != curr)
pre = pre->right;
if (pre->right == NULL) {
pre->right = curr;
curr = curr->left;
}
else {
pre->right = NULL;
int leftHeight = 0, rightHeight = 0;
Node* temp = curr->left;
while (temp != NULL) {
leftHeight++;
temp = temp->right;
}
temp = curr->right;
while (temp != NULL) {
rightHeight++;
temp = temp->left;
}
ans = max(ans,
leftHeight + rightHeight + 1);
curr = curr->right;
}
}
}
return ans;
}
// Driver code
int main()
{
// Create the given binary tree
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
// Find the diameter of the binary tree using Morris
// traversal
int diameter = findDiameter(root);
// Print the diameter of the binary tree
cout << "The diameter of given binary tree is "
<< diameter << endl;
return 0;
}
// This code is contributed by Veerendra_Singh_Rajpoot
// This code is modified by Susobhan Akhuli
//GFG
// Java code for this approach
class Node {
int data;
Node left, right;
public Node(int data) {
this.data = data;
this.left = null;
this.right = null;
}
}
public class Main
{
// Create a new node
public static Node newNode(int data) {
Node node = new Node(data);
return node;
}
// Morris traversal to find the diameter of the binary tree
public static int findDiameter(Node root) {
if(root==null)
return 0;
if(root.left==null && root.right==null)
return 1;
int ans = 0;
Node curr = root;
while (curr != null) {
if (curr.left == null) {
curr = curr.right;
} else {
Node pre = curr.left;
while (pre.right != null && pre.right != curr) {
pre = pre.right;
}
if (pre.right == null) {
pre.right = curr;
curr = curr.left;
} else {
pre.right = null;
int leftHeight = 0, rightHeight = 0;
Node temp = curr.left;
while (temp != null) {
leftHeight++;
temp = temp.right;
}
temp = curr.right;
while (temp != null) {
rightHeight++;
temp = temp.left;
}
ans = Math.max(ans, leftHeight + rightHeight + 1);
curr = curr.right;
}
}
}
return ans;
}
// Driver code to test above function
public static void main(String[] args)
{
// Create the given binary tree
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
// Find the diameter of the binary tree using Morris traversal
int diameter = findDiameter(root);
// Print the diameter of the binary tree
System.out.println("The diameter of given binary tree is " + diameter);
}
}
// This code is written by Sundaram
// This code is modified by Susobhan Akhuli
# Python3 code to implement the above approach that uses the
# morris traversal algorithm
# A tree node
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Create a new node
def newNode(data):
node = Node(data)
return node
# Morris traversal to find the diameter of the binary tree
def findDiameter(root):
if root is None:
return 0
if root.left is None and root.right is None:
return 1
ans = 0
curr = root
while curr is not None:
if curr.left is None:
curr = curr.right
else:
pre = curr.left
while pre.right is not None and pre.right != curr:
pre = pre.right
if pre.right is None:
pre.right = curr
curr = curr.left
else:
pre.right = None
leftHeight = 0
rightHeight = 0
temp = curr.left
while temp is not None:
leftHeight += 1
temp = temp.right
temp = curr.right
while temp is not None:
rightHeight += 1
temp = temp.left
ans = max(ans, leftHeight + rightHeight + 1)
curr = curr.right
return ans
# Driver code
if __name__ == '__main__':
# Create the given binary tree
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.left.right = newNode(5)
# Find the diameter of the binary tree using Morris
# traversal
diameter = findDiameter(root)
# Print the diameter of the binary tree
print("The diameter of given binary tree is", diameter)
# This code is modified by Susobhan Akhuli
using System;
public class Node {
public int data;
public Node left;
public Node right;
public Node(int data)
{
this.data = data;
this.left = null;
this.right = null;
}
}
public class Program {
public static Node newNode(int data)
{
Node node = new Node(data);
return node;
}
public static int findDiameter(Node root)
{
if(root == null)
return 0;
if(root.left == null && root.right == null)
return 1;
int ans = 0;
Node curr = root;
while (curr != null) {
if (curr.left == null) {
curr = curr.right;
}
else {
Node pre = curr.left;
while (pre.right != null
&& pre.right != curr) {
pre = pre.right;
}
if (pre.right == null) {
pre.right = curr;
curr = curr.left;
}
else {
pre.right = null;
int leftHeight = 0, rightHeight = 0;
Node temp = curr.left;
while (temp != null) {
leftHeight++;
temp = temp.right;
}
temp = curr.right;
while (temp != null) {
rightHeight++;
temp = temp.left;
}
ans = Math.Max(
ans, leftHeight + rightHeight + 1);
curr = curr.right;
}
}
}
return ans;
}
public static void Main()
{
// Create the given binary tree
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
// Find the diameter of the binary tree using Morris
// traversal
int diameter = findDiameter(root);
// Print the diameter of the binary tree
Console.WriteLine(
"The diameter of given binary tree is "
+ diameter);
}
}
// This code is modified by Susobhan Akhuli
// A tree node
class Node {
constructor(data) {
this.data = data;
this.left = null;
this.right = null;
}
}
// Create a new node
function newNode(data) {
let node = new Node(data);
return node;
}
// Morris traversal to find the diameter of the binary tree
function findDiameter(root) {
if (root == null){
return 0;
}
if(root.left == null && root.right == null){
return 1;
}
let ans = 0;
let curr = root;
while (curr != null) {
if (curr.left == null) {
curr = curr.right;
} else {
let pre = curr.left;
while (pre.right != null && pre.right != curr) {
pre = pre.right;
}
if (pre.right == null) {
pre.right = curr;
curr = curr.left;
} else {
pre.right = null;
let leftHeight = 0, rightHeight = 0;
let temp = curr.left;
while (temp != null) {
leftHeight++;
temp = temp.right;
}
temp = curr.right;
while (temp != null) {
rightHeight++;
temp = temp.left;
}
ans = Math.max(ans, leftHeight + rightHeight + 1);
curr = curr.right;
}
}
}
return ans;
}
// Driver code to test above function
// Create the given binary tree
let root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
// Find the diameter of the binary tree using Morris traversal
let diameter = findDiameter(root);
// Print the diameter of the binary tree
console.log("The diameter of given binary tree is " + diameter);
// This code is modified by Susobhan Akhuli
OutputThe diameter of given binary tree is 4
Time Complexity: O(N), where N is the number of nodes in the binary tree
Auxiliary Space: O(h), The Morris Traversal approach does not use any additional data structures like stacks or queues, which leads to an auxiliary space complexity of O(1). However, the recursion stack used by the program contributes to a space complexity of O(h), where h is the height of the binary tree.
Related Article:
Diameter of a Binary Tree in O(n) [A new method]
Diameter of an N-ary tree
Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.
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