Given a binary tree, find height of it. Height of empty tree is 0 and height of below tree is 2.
Example Tree
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 0
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.
# 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 Nikhil Kumar Singh(nickzuck_007)
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
</script>
Output
Height of tree is 2
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:**Python3
#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);
}
Time Complexity: O(n)
Space Complexity: O(n)
Time Complexity: O(n) (Please see our post for details)
