Count the number of visible nodes in Binary Tree
Last Updated :
23 Jun, 2021
Given a Binary tree, the task is to find the number of visible nodes in the given binary tree. A node is a visible node if, in the path from the root to the node N, there is no node with greater value than N’s,
Examples:
Input:
5
/ \
3 10
/ \ /
20 21 1
Output: 4
Explanation:
There are 4 visible nodes.
They are:
5: In the path 5 -> 3, 5 is the highest node value.
20: In the path 5 -> 3 -> 20, 20 is the highest node value.
21: In the path 5 -> 3 -> 21, 21 is the highest node value.
10: In the path 5 -> 10 -> 1, 10 is the highest node value.
Input:
-1
\
-2
\
-3
Output: 1
Approach: The idea is to first traverse the tree. Since we need to see the maximum value in the given path, the pre-order traversal is used to traverse the given binary tree. While traversing the tree, we need to keep the track of the maximum value of the node that we have seen so far. If the current node is greater or equal to the max value, then increment the count of the visible node and update the max value with the current node value.
Below is the implementation of the above approach:
C++
// C++ implementation to
// count the number of
// visible nodes in the
// binary tree
#include <bits/stdc++.h>
using namespace std;
// Node containing the
// left and right child of
// current node and the key value
struct Node
{
int data;
Node *left, *right;
};
/* Utility that allocates a
new node with the given key and
NULL left and right pointers. */
struct Node* newnode(int data)
{
struct Node* node = new (struct Node);
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Variable to keep the track
// of visible nodes
int countNode = 0;
// Function to perform the preorder traversal
// for the given tree
void preOrder(Node* node, int mx)
{
// Base case
if (node == NULL)
{
return;
}
// If the current node value is greater
// or equal to the max value,
// then update count variable
// and also update max variable
if (node->data >= mx)
{
countNode++;
mx = max(node->data, mx);
}
// Traverse to the left
preOrder(node->left, mx);
// Traverse to the right
preOrder(node->right, mx);
}
// Driver code
int main()
{
struct Node* root = newnode(5);
/*
5
/ \
3 10
/ \ /
20 21 1
*/
root->left = newnode(3);
root->right = newnode(10);
root->left->left = newnode(20);
root->left->right = newnode(21);
root->right->left = newnode(1);
preOrder(root, INT_MIN);
cout << countNode;
}
// This code is contributed by gauravrajput1
// Java implementation to count the
// number of visible nodes in
// the binary tree
// Class containing the left and right
// child of current node and the
// key value
class Node {
int data;
Node left, right;
// Constructor of the class
public Node(int item)
{
data = item;
left = right = null;
}
}
public class GFG {
Node root;
// Variable to keep the track
// of visible nodes
static int count;
// Function to perform the preorder traversal
// for the given tree
static void preOrder(Node node, int max)
{
// Base case
if (node == null) {
return;
}
// If the current node value is greater
// or equal to the max value,
// then update count variable
// and also update max variable
if (node.data >= max) {
count++;
max = Math.max(node.data, max);
}
// Traverse to the left
preOrder(node.left, max);
// Traverse to the right
preOrder(node.right, max);
}
// Driver code
public static void main(String[] args)
{
GFG tree = new GFG();
/*
5
/ \
3 10
/ \ /
20 21 1
*/
tree.root = new Node(5);
tree.root.left = new Node(3);
tree.root.right = new Node(10);
tree.root.left.left = new Node(20);
tree.root.left.right = new Node(21);
tree.root.right.left = new Node(1);
count = 0;
preOrder(tree.root, Integer.MIN_VALUE);
System.out.println(count);
}
}
# Python 3 implementation to
# count the number of
# visible nodes in the
# binary tree
import sys
# Node containing the
# left and right child of
# current node and the key value
class newNode:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
''' Utility that allocates a
new node with the given key and
None left and right pointers. */
'''
# Variable to keep the track
# of visible nodes
countNode = 0
# Function to perform the
# preorder traversal for the
# given tree
def preOrder(node, mx):
global countNode
# Base case
if (node == None):
return
# If the current node value
# is greater or equal to the
# max value, then update count
# variable and also update max
# variable
if (node.data >= mx):
countNode += 1
mx = max(node.data, mx)
# Traverse to the left
preOrder(node.left, mx)
# Traverse to the right
preOrder(node.right, mx)
# Driver code
if __name__ == '__main__':
root = newNode(5)
''' /*
5
/ \
3 10
/ / /
20 21 1
*/
'''
root.left = newNode(3)
root.right = newNode(10)
root.left.left = newNode(20)
root.left.right = newNode(21)
root.right.left = newNode(1)
preOrder(root, -sys.maxsize-1)
print(countNode)
# This code is contributed by SURENDRA_GANGWAR
// C# implementation to count the
// number of visible nodes in
// the binary tree
using System;
// Class containing the left and right
// child of current node and the
// key value
class Node
{
public int data;
public Node left, right;
// Constructor of the class
public Node(int item)
{
data = item;
left = right = null;
}
}
class GFG{
Node root;
// Variable to keep the track
// of visible nodes
static int count;
// Function to perform the preorder
// traversal for the given tree
static void preOrder(Node node, int max)
{
// Base case
if (node == null)
{
return;
}
// If the current node value is greater
// or equal to the max value,
// then update count variable
// and also update max variable
if (node.data >= max)
{
count++;
max = Math.Max(node.data, max);
}
// Traverse to the left
preOrder(node.left, max);
// Traverse to the right
preOrder(node.right, max);
}
// Driver code
static public void Main(String[] args)
{
GFG tree = new GFG();
/*
5
/ \
3 10
/ \ /
20 21 1
*/
tree.root = new Node(5);
tree.root.left = new Node(3);
tree.root.right = new Node(10);
tree.root.left.left = new Node(20);
tree.root.left.right = new Node(21);
tree.root.right.left = new Node(1);
count = 0;
preOrder(tree.root, int.MinValue);
Console.WriteLine(count);
}
}
// This code is contributed by Amit Katiyar
<script>
// Javascript implementation to count the
// number of visible nodes in the binary tree
// Class containing the left and right
// child of current node and the
// key value
class Node
{
constructor(item) {
this.left = null;
this.right = null;
this.data = item;
}
}
let root;
// Variable to keep the track
// of visible nodes
let count;
// Function to perform the preorder traversal
// for the given tree
function preOrder(node, max)
{
// Base case
if (node == null) {
return;
}
// If the current node value is greater
// or equal to the max value,
// then update count variable
// and also update max variable
if (node.data >= max) {
count++;
max = Math.max(node.data, max);
}
// Traverse to the left
preOrder(node.left, max);
// Traverse to the right
preOrder(node.right, max);
}
/*
5
/ \
3 10
/ \ /
20 21 1
*/
root = new Node(5);
root.left = new Node(3);
root.right = new Node(10);
root.left.left = new Node(20);
root.left.right = new Node(21);
root.right.left = new Node(1);
count = 0;
preOrder(root, Number.MIN_VALUE);
document.write(count);
// This code is contributed by rameshtravel07.
</script>
Complexity Analysis:
Time Complexity: O(N) where N is a number of nodes in the Binary tree.
Auxiliary Space: O(H) where H is the height of the Binary tree.
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