Write a Program to Find the Maximum Depth or Height of a Tree

Given a binary tree, find height of it. Height of empty tree is -1, height of tree with one node is 0 and height of below tree is 2. 

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

 maxDepth()
1. If tree is empty then return -1
2. Else
     (a) Get the max depth of left subtree recursively  i.e., 
          call maxDepth( tree->left-subtree)
     (a) Get the max depth of right subtree recursively  i.e., 
          call maxDepth( tree->right-subtree)
     (c) Get the max of max depths of left and right 
          subtrees and add 1 to it for the current node.
         max_depth = max(max dept of left subtree,  
                             max depth of right subtree) 
                             + 1
     (d) Return max_depth

See the below diagram for more clarity about execution of the recursive function maxDepth() for above example tree.

            maxDepth('1') = max(maxDepth('2'), maxDepth('3')) + 1
                               = 1 + 1
                                  /    \
                                /         \
                              /             \
                            /                 \
                          /                     \
               maxDepth('2') = 1                maxDepth('3') = 0
= max(maxDepth('4'), maxDepth('5')) + 1
= 1 + 0   = 1         
                   /    \
                 /        \
               /            \
             /                \
           /                    \
 maxDepth('4') = 0     maxDepth('5') = 0

Implementation:

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 our post Tree Traversal for details)

Auxiliary Space: O(n) due to recursive calls.

Method 2: Another method to solve this problem is to do Level Order Traversal. While doing the 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.

Implementation:

C++

#include <iostream>
#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(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(12);
    root->right = newNode(13);

    root->right->left   = newNode(14);
    root->right->right  = newNode(15);

    root->right->left->left   = newNode(21);
    root->right->left->right  = newNode(22);
    root->right->right->left  = newNode(23);
    root->right->right->right = newNode(24);

      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(12);
    root.right = newNode(13);

    root.right.left   = newNode(14);
    root.right.right  = newNode(15);

    root.right.left.left   = newNode(21);
    root.right.left.right  = newNode(22);
    root.right.right.left  = newNode(23);
    root.right.right.right = newNode(24);

    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(12)
root.right = newNode(13)

root.right.left = newNode(14)
root.right.right = newNode(15)

root.right.left.left = newNode(21)
root.right.left.right = newNode(22)
root.right.right.left = newNode(23)
root.right.right.right = newNode(24)

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


# This code is contributed by shinjanpatra

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(12)
root.right = newNode(13)

root.right.left = newNode(14)
root.right.right = newNode(15)

root.right.left.left = newNode(21)
root.right.left.right = newNode(22)
root.right.right.left = newNode(23)
root.right.right.right = newNode(24)

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

// This code is contributed by shinjanpatra

</script>

Output

Height(Depth) of tree is: 4

Time Complexity: O(n)

Auxiliary Space: O(n)