Find the Maximum Depth or Height of given Binary Tree

Given a binary tree, the task is to find the height of the tree. The height of the tree is the number of edges in the tree from the root to the deepest node. 

Note: The height of an empty tree is 0.

Example of Binary Tree

Recursively calculate the height of the left and the right subtrees of a node and assign height to the node as max of the heights of two children plus 1. See below the pseudo code and program for details.

Illustration:

Consider the following tree:

Example of Tree

maxDepth(‘1’) = max(maxDepth(‘2’), maxDepth(‘3’)) + 1 = 2 + 1

because recursively 
maxDepth(‘2’) =  max (maxDepth(‘4’), maxDepth(‘5’)) + 1 = 1 + 1 and  (as height of both ‘4’ and ‘5’ are 1)
maxDepth(‘3’) = 1

Follow the below steps to Implement the idea:

• Recursively do a Depth-first search.
• If the tree is empty then return 0
• Otherwise, do the following
  • Get the max depth of the left subtree recursively  i.e. call maxDepth( tree->left-subtree)
  • Get the max depth of the right subtree recursively  i.e. call maxDepth( tree->right-subtree)
  • Get the max of max depths of left and right subtrees and add 1 to it for the current node.
    • 
• Return max_depth.

Below is the Implementation of the above approach:

C++

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

/* A binary tree node has data, pointer to left child
and a pointer to right child */
class node {
public:
    int data;
    node* left;
    node* right;
};

/* Compute the "maxDepth" of a tree -- the number of
    nodes along the longest path from the root node
    down to the farthest leaf node.*/
int maxDepth(node* node)
{
    if (node == NULL)
        return 0;
    else {
        /* compute the depth of each subtree */
        int lDepth = maxDepth(node->left);
        int rDepth = maxDepth(node->right);

        /* use the larger one */
        if (lDepth > rDepth)
            return (lDepth + 1);
        else
            return (rDepth + 1);
    }
}

/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
node* newNode(int data)
{
    node* Node = new node();
    Node->data = data;
    Node->left = NULL;
    Node->right = NULL;

    return (Node);
}

// Driver code
int main()
{
    node* root = newNode(1);

    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);

    cout << "Height of tree is " << maxDepth(root);
    return 0;
}

// This code is contributed by Amit Srivastav

C

#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;
    struct node* right;
};

/* Compute the "maxDepth" of a tree -- the number of
    nodes along the longest path from the root node
    down to the farthest leaf node.*/
int maxDepth(struct node* node)
{
    if (node == NULL)
        return 0;
    else {
        /* compute the depth of each subtree */
        int lDepth = maxDepth(node->left);
        int rDepth = maxDepth(node->right);

        /* use the larger one */
        if (lDepth > rDepth)
            return (lDepth + 1);
        else
            return (rDepth + 1);
    }
}

/* 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);
}

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("Height of tree is %d", maxDepth(root));

    getchar();
    return 0;
}

Java

// Java program to find height of tree

// A binary tree node
class Node {
    int data;
    Node left, right;

    Node(int item)
    {
        data = item;
        left = right = null;
    }
}

class BinaryTree {
    Node root;

    /* Compute the "maxDepth" of a tree -- the number of
       nodes along the longest path from the root node
       down to the farthest leaf node.*/
    int maxDepth(Node node)
    {
        if (node == null)
            return 0;
        else {
            /* compute the depth of each subtree */
            int lDepth = maxDepth(node.left);
            int rDepth = maxDepth(node.right);

            /* use the larger one */
            if (lDepth > rDepth)
                return (lDepth + 1);
            else
                return (rDepth + 1);
        }
    }

    /* Driver program to test above functions */
    public static void main(String[] args)
    {
        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);

        System.out.println("Height of tree is "
                           + tree.maxDepth(tree.root));
    }
}

// This code has been contributed by Amit Srivastav

Python3

# Python3 program to find the maximum depth of 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

# Compute the "maxDepth" of a tree -- the number of nodes
# along the longest path from the root node down to the
# farthest leaf node


def maxDepth(node):
    if node is None:
        return 0

    else:

        # Compute the depth of each subtree
        lDepth = maxDepth(node.left)
        rDepth = maxDepth(node.right)

        # Use the larger one
        if (lDepth > rDepth):
            return lDepth+1
        else:
            return rDepth+1


# Driver program to test above function
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)


print("Height of tree is %d" % (maxDepth(root)))

# This code is contributed by Amit Srivastav

C#

// C# program to find height of tree
using System;

// A binary tree node
public class Node {
    public int data;
    public Node left, right;

    public Node(int item)
    {
        data = item;
        left = right = null;
    }
}

public class BinaryTree {
    Node root;

    /* Compute the "maxDepth" of a tree -- the number of
    nodes along the longest path from the root node
    down to the farthest leaf node.*/
    int maxDepth(Node node)
    {
        if (node == null)
            return 0;
        else {
            /* compute the depth of each subtree */
            int lDepth = maxDepth(node.left);
            int rDepth = maxDepth(node.right);

            /* use the larger one */
            if (lDepth > rDepth)
                return (lDepth + 1);
            else
                return (rDepth + 1);
        }
    }

    /* Driver code */
    public static void Main(String[] args)
    {
        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);

        Console.WriteLine("Height of tree is "
                          + tree.maxDepth(tree.root));
    }
}

// This code has been contributed by
// Correction done by Amit Srivastav

Javascript

<script>

// JavaScript program to find height of tree

// A binary tree node
class Node
{
    constructor(item)
    {
        this.data=item;
        this.left=this.right=null;
    }
}

    let root;

     /* Compute the "maxDepth" of a tree -- the number of
       nodes along the longest path from the root node
       down to the farthest leaf node.*/
    function maxDepth(node)
    {
        if (node == null)
            return 0;
        else
        {
            /* compute the depth of each subtree */
            let lDepth = maxDepth(node.left);
            let rDepth = maxDepth(node.right);

            /* use the larger one */
            if (lDepth > rDepth)
                return (lDepth + 1);
             else
                return (rDepth + 1);
        }
    }

    /* Driver program to test above functions */

        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);

        document.write("Height of tree is : " +
                                      maxDepth(root));




// This code is contributed by rag2127
//Correction done by Amit Srivastav

</script>

Output

Height of tree is 3

Time Complexity: O(N) (Please see the post on Tree Traversal for details)
Auxiliary Space: O(N) due to recursive stack.

Find the Maximum Depth or Height of a Tree using Level Order Traversal:

Do Level Order Traversal, while adding Nodes at each level to Queue, we have to add NULL Node so that whenever it is encountered, we can increment the value of variable and that level get counted.

Follow the below steps to Implement the idea:

• Traverse the tree in level order traversal starting from root.
  • Initialize an empty queue Q, a variable depth and push root, then push null into the Q.
  • Run a while loop till Q is not empty.
    • Store the front element of Q and Pop out the front element.
    • If the front of Q is NULL then increment depth by one and if queue is not empty then push NULL into the Q.
    • Else if the element is not NULL then check for its left and right children and if they are not NULL push them into Q.
• Return depth.

Below is the Implementation of the above approach:

C++

#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// A Tree node
struct Node {
    int key;
    struct Node *left, *right;
};

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

/*Function to find the height(depth) of the tree*/
int height(struct Node* root)
{

    // Initialising a variable to count the
    // height of tree
    int depth = 0;

    queue<Node*> q;

    // Pushing first level element along with NULL
    q.push(root);
    q.push(NULL);
    while (!q.empty()) {
        Node* temp = q.front();
        q.pop();

        // When NULL encountered, increment the value
        if (temp == NULL) {
            depth++;
        }

        // If NULL not encountered, keep moving
        if (temp != NULL) {
            if (temp->left) {
                q.push(temp->left);
            }
            if (temp->right) {
                q.push(temp->right);
            }
        }

        // If queue still have elements left,
        // push NULL again to the queue.
        else if (!q.empty()) {
            q.push(NULL);
        }
    }
    return depth;
}

// Driver program
int main()
{
    // Let us create Binary Tree shown in above example
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);

    root->left->left = newNode(4);
    root->left->right = newNode(5);

    cout << "Height(Depth) of tree is: " << height(root);
}

Java

// Java program for above approach
import java.util.LinkedList;
import java.util.Queue;

class GFG {

    // A tree node structure
    static class Node {
        int key;
        Node left;
        Node right;
    }

    // Utility function to create
    // a new node
    static Node newNode(int key)
    {
        Node temp = new Node();
        temp.key = key;
        temp.left = temp.right = null;
        return temp;
    }

    /*Function to find the height(depth) of the tree*/
    public static int height(Node root)
    {

        // Initialising a variable to count the
        // height of tree
        int depth = 0;

        Queue<Node> q = new LinkedList<>();

        // Pushing first level element along with null
        q.add(root);
        q.add(null);
        while (!q.isEmpty()) {
            Node temp = q.peek();
            q.remove();

            // When null encountered, increment the value
            if (temp == null) {
                depth++;
            }

            // If null not encountered, keep moving
            if (temp != null) {
                if (temp.left != null) {
                    q.add(temp.left);
                }
                if (temp.right != null) {
                    q.add(temp.right);
                }
            }

            // If queue still have elements left,
            // push null again to the queue.
            else if (!q.isEmpty()) {
                q.add(null);
            }
        }
        return depth;
    }

    // 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("Height(Depth) of tree is: "
                           + height(root));
    }
}

// This code is contributed by jana_sayantan.

Python3

# Python code to implement the approach

# A Tree node


class Node:

    def __init__(self):
        self.key = 0
        self.left, self.right = None, None

# Utility function to create a new node


def newNode(key):

    temp = Node()
    temp.key = key
    temp.left, temp.right = None, None
    return temp


# Function to find the height(depth) of the tree
def height(root):

    # Initialising a variable to count the
    # height of tree
    depth = 0

    q = []

    # appending first level element along with None
    q.append(root)
    q.append(None)
    while(len(q) > 0):
        temp = q[0]
        q = q[1:]

        # When None encountered, increment the value
        if(temp == None):
            depth += 1

        # If None not encountered, keep moving
        if(temp != None):
            if(temp.left):
                q.append(temp.left)

            if(temp.right):
                q.append(temp.right)

        # If queue still have elements left,
        # append None again to the queue.
        elif(len(q) > 0):
            q.append(None)
    return depth

# Driver program


# Let us create Binary Tree shown in above example
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)

root.left.left = newNode(4)
root.left.right = newNode(5)

print(f"Height(Depth) of tree is: {height(root)}")


# This code is contributed by shinjanpatra

C#

// C# Program to find the Maximum Depth or Height of Binary Tree
using System;
using System.Collections.Generic;

// A Tree node
public class Node {
    public int data;
    public Node left, right;

    public Node(int item)
    {
        data = item;
        left = null;
        right = null;
    }
}

public class BinaryTree {

    Node root;

    // Function to find the height(depth) of the tree
    int height()
    {
        // Initialising a variable to count the
        // height of tree
        int depth = 0;

        Queue<Node> q = new Queue<Node>();

        // Pushing first level element along with NULL
        q.Enqueue(root);
        q.Enqueue(null);
        while (q.Count != 0) {
            Node temp = q.Dequeue();

            // When NULL encountered, increment the value
            if (temp == null)
                depth++;

            // If NULL  not encountered, keep moving
            if (temp != null) {
                if (temp.left != null) {
                    q.Enqueue(temp.left);
                }
                if (temp.right != null) {
                    q.Enqueue(temp.right);
                }
            }

            // If queue still have elements left,
            // push NULL again to the queue
            else if (q.Count != 0) {
                q.Enqueue(null);
            }
        }
        return depth;
    }

    // Driver program
    public static void Main()
    {
        // Let us create Binary Tree shown in above example
        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);

        Console.WriteLine("Height(Depth) of tree is: "
                          + tree.height());
    }
}

// This code is contributed by Yash Agarwal(yashagarwal2852002)

Javascript

<script>

// JavaScript code to implement the approach

// A Tree node
class Node{

    constructor(){
        this.key = 0
        this.left = null
        this.right = null
    }

}

// Utility function to create a new node
function newNode(key){

    let temp = new Node()
    temp.key = key
    temp.left = null
    temp.right = null
    return temp

}

// Function to find the height(depth) of the tree
function height(root){

    // Initialising a variable to count the
    // height of tree
    let depth = 0

    let q = []

    // pushing first level element along with null
    q.push(root)
    q.push(null)
    while(q.length>0){
        let temp = q.shift()

        // When null encountered, increment the value
        if(temp == null)
            depth += 1

        // If null not encountered, keep moving
        if(temp != null){
            if(temp.left)
                q.push(temp.left)

            if(temp.right)
                q.push(temp.right)
        }

        // If queue still have elements left,
        // push null again to the queue.
        else if(q.length>0)
            q.push(null)
    }
    return depth

}

// Driver program

// Let us create Binary Tree shown in above example
let root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)

root.left.left = newNode(4)
root.left.right = newNode(5)

document.write(`Height(Depth) of tree is: ${height(root)}`,"</br>")

// This code is contributed by shinjanpatra

</script>

Output

Height(Depth) of tree is: 3

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

Another method to find height using Level Order Traversal:

This method also uses the concept of Level Order Traversal but we wont be adding null in the Queue. Simply increase the counter when the level increases and push the children of current node into the queue, then remove all the nodes from the queue of the current Level.

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// A Tree node
struct Node {
  int key;
  struct Node *left, *right;
};

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

/*Function to find the height(depth) of the tree*/
int height(Node* root)
{

  // Initialising a variable to count the
  // height of tree
  queue<Node*> q;
  q.push(root);
  int height = 0;
  while (!q.empty()) {
    int size = q.size();
    for (int i = 0; i < size; i++) {
      Node* temp = q.front();
      q.pop();
      if (temp->left != NULL) {
        q.push(temp->left);
      }
      if (temp->right != NULL) {
        q.push(temp->right);
      }
    }
    height++;
  }
  return height;
}

// Driver program
int main()
{

  // Let us create Binary Tree shown in above example
  Node* root = newNode(1);
  root->left = newNode(2);
  root->right = newNode(3);

  root->left->left = newNode(4);
  root->left->right = newNode(5);

  cout << "Height(Depth) of tree is: " << height(root);
}

// This code is contributed by Abhijeet Kumar(abhijeet19403)

Java

// Java program for above approach
import java.util.LinkedList;
import java.util.Queue;

class GFG {

    // A tree node structure
    static class Node {
        int key;
        Node left;
        Node right;
    }

    // Utility function to create
    // a new node
    static Node newNode(int key)
    {
        Node temp = new Node();
        temp.key = key;
        temp.left = temp.right = null;
        return temp;
    }

    /*Function to find the height(depth) of the tree*/
    public static int height(Node root)
    {

        // Initialising a variable to count the
        // height of tree
        Queue<Node> q = new LinkedList<Node>();
        q.add(root);
        int height = 0;
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                Node temp = q.poll();
                if (temp.left != null) {
                    q.add(temp.left);
                }
                if (temp.right != null) {
                    q.add(temp.right);
                }
            }
            height++;
        }
        return height;
    }

    // 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("Height(Depth) of tree is: "
                           + height(root));
    }
}

C#

using System;
using System.Collections.Generic;

class GFG {

  // A Tree node
  class Node {
    public int key;
    public Node left, right;

    public Node(int key)
    {
      this.key=key;
      this.left=this.right=null;
    }
  }

  // Utility function to create a new node
  /*Node newNode(int key)
    {
      Node* temp = new Node;
      temp.key = key;
      temp.left = temp.right = NULL;
      return (temp);
    }*/

  /*Function to find the height(depth) of the tree*/
  static int height(Node root)
  {

    // Initialising a variable to count the
    // height of tree
    Queue<Node> q=new Queue<Node>();
    q.Enqueue(root);
    int height = 0;
    while (q.Count>0) {
      int size = q.Count;
      for (int i = 0; i < size; i++) {
        Node temp = q.Peek();
        q.Dequeue();
        if (temp.left != null) {
          q.Enqueue(temp.left);
        }
        if (temp.right != null) {
          q.Enqueue(temp.right);
        }
      }
      height++;
    }
    return height;
  }

  // Driver program
  public static void Main()
  {

    // Let us create Binary Tree shown in above example
    Node 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);

    Console.Write("Height(Depth) of tree is: " + height(root));
  }
}

// This code is contributed by poojaagarwal2.

Python3

# Python3 program to find the height of a tree

# A binary tree node
class Node:

    # Constructor to create a new node
    def __init__(self, data):
        self.key = data
        self.left = None
        self.right = None

# Function to find height of tree
def height(root):
    # Base Case
    if root is None:
        return 0

    # Create an empty queue for level order traversal
    q = []

    # Enqueue Root and initialize height
    q.append(root)
    height = 0

    # Loop while queue is not empty
    while q:

        # nodeCount (queue size) indicates number of nodes
        # at current level
        nodeCount = len(q)

        # Dequeue all nodes of current level and Enqueue all
        # nodes of next level
        while nodeCount > 0:
            node = q.pop(0)
            if node.left is not None:
                q.append(node.left)
            if node.right is not None:
                q.append(node.right)
            nodeCount -= 1
        height += 1

    return height

# Driver Code
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)

print("Height(Depth) of tree is", height(root))

Javascript

// JavaScript program for above approach
// a tree node
class Node{
    constructor(key){
        this.key = key;
        this.left = this.right = null;
    }
}

// utility function to create a new node
function newNode(key){
    return new Node(key);
}

// function to find the  height of the tree
function height(root){
    // initialising a variable to count the
    // height of tree
    let q = [];
    q.push(root);
    let height = 0;
    while(q.length > 0){
        let size = q.length;
        for(let i = 0; i<size; i++){
            let temp =  q.shift();
            if(temp.left != null){
                q.push(temp.left);
            }
            if(temp.right != null){
                q.push(temp.right);
            }
        }
        height++;
    }
    return height;
}

// driver code
let root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);

document.write("Height(Depth) of tree is: " + height(root));

// this code is contributed by Kirti Agarwal(kirtiagarwal23121999)

Output

Height(Depth) of tree is: 3

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

Finding height using Morris Traversal Algorithm:

The basic idea behind the algorithm is to traverse the left subtree of each node first and then move to its right subtree. During the traversal, we keep track of the maximum depth of the tree by incrementing a counter variable each time we move down a level in the tree.

Follow the below steps to implement the above idea:

• Initialize the current node as the root of the tree.
• While the current node is not NULL:
  • If the left child of the current node is NULL, visit the current node and move to its 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 this node is NULL, set it to the current node and move to the left child of the current node.
    • If the right child of this node is not NULL, set it back to NULL, visit the current node, and move to its right child.
• Return the maximum depth of the tree.

Below is the implementation of the above approach:

C++

// C++ code to implement the above approach using morris
// traversal Algorithm
#include <iostream>
using namespace std;

// Definition of a binary tree node
struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x)
        : val(x)
        , left(NULL)
        , right(NULL)
    {
    }
};

// Function to find the height of a binary tree using Morris
// Traversal technique
int findHeight(TreeNode* root)
{
    int height = 0;
    TreeNode* current = root;
    while (current != NULL) {
        if (current->left == NULL) {
            // If left subtree is NULL, move to right
            // subtree
            current = current->right;
            height++; // Increment the height of the tree
        }
        else {
            // Find the inorder predecessor of current node
            TreeNode* pre = current->left;
            while (pre->right != NULL
                   && pre->right != current)
                pre = pre->right;

            if (pre->right == NULL) {
                // Make current node the right child of its
                // inorder predecessor
                pre->right = current;
                current = current->left;
            }
            else {
                // If the right child of the inorder
                // predecessor already points to the current
                // node, then we have traversed the left
                // subtree and its inorder traversal is
                // complete.
                pre->right = NULL;
                current = current->right; // Move to the
                                          // right subtree
            }
        }
    }
    return height;
}

// Driver Code
int main()
{
    // Create the binary tree
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);

    // Calculate the height of the tree using Morris
    // Traversal
    int height = findHeight(root);

    // Output the height of the tree
    cout << "Height of the binary tree is: " << height
         << endl;

    return 0;
}
// This Code is Contributed by Veerendra_Singh_Rajpoot

Java

// Java code to implement the above approach using morris
// traversal Algorithm
import java.util.*;

// Definition of a binary tree node
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x)
    {
        val = x;
        left = null;
        right = null;
    }
};

public class Main {
    // Function to find the height of a binary tree using
    // Morris Traversal technique
    public static int findHeight(TreeNode root)
    {
        int height = 0;
        TreeNode current = root;
        while (current != null) {
            if (current.left == null) {
                // If left subtree is null, move to right
                // subtree
                current = current.right;
                height++; // Increment the height of the
                          // tree
            }
            else {
                // Find the inorder predecessor of current
                // node
                TreeNode pre = current.left;
                while (pre.right != null
                       && pre.right != current)
                    pre = pre.right;

                if (pre.right == null) {
                    // Make current node the right child of
                    // its inorder predecessor
                    pre.right = current;
                    current = current.left;
                }
                else {
                    // If the right child of the inorder
                    // predecessor already points to the
                    // current node, then we have traversed
                    // the left subtree and its inorder
                    // traversal is complete.
                    pre.right = null;
                    current
                        = current.right; // Move to the
                                         // right subtree
                }
            }
        }
        return height;
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Create the binary tree
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);

        // Calculate the height of the tree using Morris
        // Traversal
        int height = findHeight(root);

        // Output the height of the tree
        System.out.println("Height of the binary tree is: "
                           + height);
    }
}
// This Code is Contributed by rutikbhosale

Python3

# Python3 code to implement the above approach using morris
# traversal Algorithm


class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

# Function to find the height of a binary tree using Morris
# Traversal technique


def findHeight(root):
    height = 0
    current = root
    while current != None:
        if current.left == None:
            # If left subtree is NULL, move to right
            # subtree
            current = current.right
            height += 1  # Increment the height of the tree
        else:
            # Find the inorder predecessor of current node
            pre = current.left
            while pre.right != None and pre.right != current:
                pre = pre.right

            if pre.right == None:
                # Make current node the right child of its
                # inorder predecessor
                pre.right = current
                current = current.left
            else:
                # If the right child of the inorder
                # predecessor already points to the current
                # node, then we have traversed the left
                # subtree and its inorder traversal is
                # complete.
                pre.right = None
                current = current.right  # Move to the
                # right subtree
    return height


# Driver Code
if __name__ == '__main__':
    # Create the binary tree
    root = TreeNode(1)
    root.left = TreeNode(2)
    root.right = TreeNode(3)
    root.left.left = TreeNode(4)
    root.left.right = TreeNode(5)

    # Calculate the height of the tree using Morris
    # Traversal technique
    print("Height of the binary tree is:", findHeight(root))

Javascript

// Definition of a binary tree node
class TreeNode {
    constructor(x) {
        this.val = x;
        this.left = null;
        this.right = null;
    }
}

// Function to find the height of
// a binary tree using Morris Traversal technique
function findHeight(root) {
    let height = 0;
    let current = root;
    while (current != null)
    {
        if (current.left == null)
        {

            // If left subtree is NULL, move to right subtree
            current = current.right;
            height++; // Increment the height of the tree
        }
        else
        {

            // Find the inorder predecessor of current node
            let pre = current.left;
            while (pre.right != null && pre.right != current)
                pre = pre.right;

            if (pre.right == null)
            {

                // Make current node the right child of its inorder predecessor
                pre.right = current;
                current = current.left;
            }
            else
            {

                // If the right child of the inorder predecessor
                // already points to the current node,
                // then we have traversed the left subtree and
                // its inorder traversal is complete.
                pre.right = null;
                current = current.right; // Move to the right subtree
            }
        }
    }
    return height;
}

// Driver Code

// Create the binary tree
let root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);

// Calculate the height of the tree using Morris Traversal
let height = findHeight(root);
console.log("Height of the binary tree is: " + height);

C#

// C# code to implement the above approach using morris
// traversal Algorithm
using System;

// Definition of a binary tree node
public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int x)
    {
        val = x;
        left = null;
        right = null;
    }
}

public class Program {
    // Function to find the height of a binary tree using
    // Morris Traversal technique
    public static int FindHeight(TreeNode root)
    {
        int height = 0;
        TreeNode current = root;
        while (current != null) {
            if (current.left == null) {
                // If left subtree is null, move to right
                // subtree
                current = current.right;
                height++; // Increment the height of the
                // tree
            }
            else {
                // Find the inorder predecessor of current
                // node
                TreeNode pre = current.left;
                while (pre.right != null
                       && pre.right != current)
                    pre = pre.right;
                if (pre.right == null) {
                    // Make current node the right child of
                    // its inorder predecessor
                    pre.right = current;
                    current = current.left;
                }
                else {
                    // If the right child of the inorder
                    // predecessor already points to the
                    // current node, then we have traversed
                    // the left subtree and its inorder
                    // traversal is complete.
                    pre.right = null;
                    current
                        = current.right; // Move to the
                                         // right subtree
                }
            }
        }
        return height;
    }

    // Driver Code
    public static void Main()
    {
        // Create the binary tree
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
        root.left.left = new TreeNode(4);
        root.left.right = new TreeNode(5);

        // Calculate the height of the tree using Morris
        // Traversal
        int height = FindHeight(root);

        // Output the height of the tree
        Console.WriteLine("Height of the binary tree is: "
                          + height);
    }
}

Output

Height of the binary tree is: 3

Time Complexity: O(N), The time complexity of finding the height of a binary tree using Morris traversal is O(n), where n is the number of nodes in the tree.

Auxiliary Space: O(1), The space complexity of the algorithm is O(1), which is constant space complexity. This is because we are not using any additional data structures to store nodes or keep track of the traversal