Given a Binary Tree consisting of N nodes, the task is to find the number of paths from the root to any node X, such that all the node values in that path are at most X.
Examples:
Input: Below is the given Tree:
Output: 4
Explanation:
The paths from the root to any node X that have value at most values of node X are:
- Node 3(root node): It always follows the given property.
- Node 4: The path starting from the root to node with value 4 has order (3 ? 4), with the maximum value of a node being 4.
- Node 5: The path starting from the root to node with value 5 has order (3 ? 4 ? 5), with the maximum value of a node being 5.
- Node 3: The path starting from the root to node with value 3 has order (3 ? 1 ? 3), with the maximum value of a node being 3.
Therefore, the count of required paths is 4.
Input: Below is the given Tree:
Output: 3
Approach – using DFS: The idea is to traverse the tree using a Depth First Search traversal while checking if the maximum value from root to any node X is equal to X or not.
Follow the steps below to solve the problem:
- Initialize a variable, say count as 0 to store the count of paths from the root to any node X having all the node values in that path is at most X.
- Traverse the tree recursively using depth-first-search and perform the following steps:
- Every recursive call for DFS Traversal, apart from the parent node, pass the maximum value of the node obtained so far in that path.
- Check if the current node value is greater or equal to the maximum value obtained so far, then increment the value of count by 1 and update the maximum value to the current node value.
- After completing the above steps, print the value of count as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Node structure of the binary treestruct Node { int val; Node *left, *right;};// Function for creating new nodestruct Node* newNode(int data){ // Allocate memory for new node struct Node* temp = new Node(); // Assigning data value temp->val = data; temp->left = NULL; temp->right = NULL; // Return the Node return temp;}// Function to perform the DFS Traversal// to find the number of paths having// root to node X has value at most Xint countNodes(Node* root, int max){ // If the root node is NULL if (!root) return 0; // Check if the current value is // greater than the maximum value // in path from root to current node if (root->val >= max) return 1 + countNodes(root->left, root->val) + countNodes(root->right, root->val); // Otherwise return countNodes(root->left, max) + countNodes(root->right, max);}// Driver Codeint main(){ // Given Binary Tree Node* root = NULL; root = newNode(3); root->left = newNode(1); root->right = newNode(4); root->left->left = newNode(3); root->right->left = newNode(1); root->right->right = newNode(5); cout << countNodes(root, INT_MIN); return 0;} |
Java
// Java program for the above approachimport java.util.*;// Class containing left and// right child of current// node and key valueclass Node { int data; Node left, right; public Node(int item) { data = item; left = right = null; }} class GFG { // Root of the Binary Tree Node root; public GFG() { root = null; } // Function to perform the DFS Traversal// to find the number of paths having// root to node X has value at most Xstatic int countNodes(Node root, int max){ // If the root node is NULL if (root == null) return 0; // Check if the current value is // greater than the maximum value // in path from root to current node if (root.data >= max) return 1 + countNodes(root.left, root.data) + countNodes(root.right, root.data); // Otherwise return countNodes(root.left, max) + countNodes(root.right, max);} // Driver codepublic static void main (String[] args) { GFG tree = new GFG(); tree.root = new Node(3); tree.root.left = new Node(1); tree.root.right = new Node(4); tree.root.left.left = new Node(3); tree.root.right.left = new Node(1); tree.root.right.right = new Node(5); System.out.println(countNodes(tree.root, Integer.MIN_VALUE)); }}// This code is contributed by offbeat |
Python3
# Python3 program for the above approach# Node structure of the binary treeclass Node: def __init__(self, x): self.val = x self.left = None self.right = None# Function to perform the DFS Traversal# to find the number of paths having# root to node X has value at most Xdef countNodes(root, max): # If the root node is NULL if (not root): return 0 # Check if the current value is # greater than the maximum value #in path from root to current node if (root.val >= max): return 1 + countNodes(root.left,root.val) + countNodes(root.right, root.val) # Otherwise return countNodes(root.left, max) + countNodes(root.right, max)# Driver Codeif __name__ == '__main__': # Given Binary Tree root = Node(3) root.left = Node(1) root.right = Node(4) root.left.left = Node(3) root.right.left = Node(1) root.right.right = Node(5) print(countNodes(root, -10**19))# This code is contributed by mohit kumar 29. |
C#
// C# program to count frequencies of array itemsusing System;// Class containing left and// right child of current// node and key valueclass Node { public int data; public Node left, right; public Node(int item) { data = item; left = right = null; }} public class GFG{ // Root of the Binary Tree Node root; public GFG() { root = null; } // Function to perform the DFS Traversal// to find the number of paths having// root to node X has value at most Xstatic int countNodes(Node root, int max){ // If the root node is NULL if (root == null) return 0; // Check if the current value is // greater than the maximum value // in path from root to current node if (root.data >= max) return 1 + countNodes(root.left, root.data) + countNodes(root.right, root.data); // Otherwise return countNodes(root.left, max) + countNodes(root.right, max);}// Driver codepublic static void Main(String []args){ GFG tree = new GFG(); tree.root = new Node(3); tree.root.left = new Node(1); tree.root.right = new Node(4); tree.root.left.left = new Node(3); tree.root.right.left = new Node(1); tree.root.right.right = new Node(5); Console.WriteLine(countNodes(tree.root, Int32.MinValue)); }}// This code is contributed by jana_sayantan. |
Javascript
<script> // Javascript program for the above approach // Class containing left and // right child of current // node and key value class Node { constructor(item) { this.left = null; this.right = null; this.data = item; } } // Root of the Binary Tree let root; class GFG { constructor() { root = null; } } // Function to perform the DFS Traversal // to find the number of paths having // root to node X has value at most X function countNodes(root, max) { // If the root node is NULL if (root == null) return 0; // Check if the current value is // greater than the maximum value // in path from root to current node if (root.data >= max) return 1 + countNodes(root.left, root.data) + countNodes(root.right, root.data); // Otherwise return countNodes(root.left, max) + countNodes(root.right, max); } let tree = new GFG(); tree.root = new Node(3); tree.root.left = new Node(1); tree.root.right = new Node(4); tree.root.left.left = new Node(3); tree.root.right.left = new Node(1); tree.root.right.right = new Node(5); document.write(countNodes(tree.root, Number.MIN_VALUE));// This code is contributed by sureh07.</script> |
Output:
4
Time Complexity: O(N)
Auxiliary Space: O(N)
Approach using BFS: The idea is to traverse the tree using breadth-first search while checking if the maximum value from root to X is equal to X. Follow the steps below to solve the problem:
- Initialize a variable, say count as 0 to store the count of paths from the root to any node X having all the node values in that path is at most X and a queue Q of pairs to perform the BFS Traversal.
- Push the root node with INT_MIN as the maximum value in the queue.
- Now, until Q is non-empty perform the following:
- Pop the front node from the queue.
- If the front node value is at least the current maximum value obtained so far, then increment the value of count by 1.
- Update the maximum value that occurred so far with the current node value.
- If the left and right nodes exist for the current popped node then push it into the queue Q with the updated maximum value in the above step.
- After completing the above steps, print the value of count as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Node of the binary treestruct Node { int val; Node *left, *right;};// Function for creating new nodestruct Node* newNode(int data){ // Allocate memory for new node struct Node* temp = new Node(); temp->val = data; temp->left = NULL; temp->right = NULL; // Return the created node return temp;}// Function to perform the DFS Traversal// to find the number of paths having// root to node X has value at most Xint countNodes(Node* root){ // Initialize queue queue<pair<Node*, int> > q; int m = INT_MIN; // Push root in queue with the // maximum value m q.push({ root, m }); // Stores the count of good nodes int count = 0; // Traverse all nodes while (!q.empty()) { // Store the front node of // the queue auto temp = q.front(); q.pop(); Node* node = temp.first; int num = temp.second; // Check is current node is // greater than the maximum // value in path from root to // the current node if (node->val >= num) count++; // Update the maximum value m m = max(node->val, num); // If left child is not null, // push it to queue with the // maximum value m if (node->left) q.push({ node->left, m }); // If right child is not null, // push it to queue with the // maximum value m if (node->right) q.push({ node->right, m }); } // Returns the answer return count;}// Driver Codeint main(){ // Construct a Binary Tree Node* root = NULL; root = newNode(3); root->left = newNode(1); root->right = newNode(4); root->left->left = newNode(3); root->right->left = newNode(1); root->right->right = newNode(5); cout << countNodes(root); return 0;} |
Java
// Java program for the above approachimport java.util.*;// Node of the binary treeclass Node { int val; Node left, right; public Node(int data) { val = data; left = right = null; }}// User defined Pair classclass Pair { Node x; int y; // Constructor public Pair(Node x, int y) { this.x = x; this.y = y; }} public class Main{ // Function for creating new node static Node newNode(int data) { // Allocate memory for new node Node temp = new Node(data); // Return the created node return temp; } // Function to perform the DFS Traversal // to find the number of paths having // root to node X has value at most X static int countNodes(Node root) { // Initialize queue Vector<Pair> q = new Vector<Pair>(); int m = Integer.MIN_VALUE; // Push root in queue with the // maximum value m q.add(new Pair(root, m)); // Stores the count of good nodes int count = 0; // Traverse all nodes while (q.size() > 0) { // Store the front node of // the queue Pair temp = q.get(0); q.remove(0); Node node = temp.x; int num = temp.y; // Check is current node is // greater than the maximum // value in path from root to // the current node if (node.val >= num) count++; // Update the maximum value m m = Math.max(node.val, num); // If left child is not null, // push it to queue with the // maximum value m if (node.left != null) q.add(new Pair(node.left, m)); // If right child is not null, // push it to queue with the // maximum value m if (node.right != null) q.add(new Pair(node.right, m)); } // Returns the answer return count; } public static void main(String[] args) { // Construct a Binary Tree Node root = null; root = newNode(3); root.left = newNode(1); root.right = newNode(4); root.left.left = newNode(3); root.right.left = newNode(1); root.right.right = newNode(5); System.out.println(countNodes(root)); }}// This code is contributed by mukesh07. |
Python3
# Python3 program for the above approachimport sys # Node of the binary treeclass Node: def __init__(self, data): self.val = data self.left = None self.right = None# Function for creating new nodedef newNode(data): # Allocate memory for new node temp = Node(data) # Return the created node return temp# Function to perform the DFS Traversal# to find the number of paths having# root to node X has value at most Xdef countNodes(root): # Initialize queue q = [] m = -sys.maxsize # Push root in queue with the # maximum value m q.append([ root, m ]) # Stores the count of good nodes count = 0 # Traverse all nodes while (len(q) > 0): # Store the front node of # the queue temp = q[0] q.pop(0) node = temp[0] num = temp[1] # Check is current node is # greater than the maximum # value in path from root to # the current node if (node.val >= num): count+=1 # Update the maximum value m m = max(node.val, num) # If left child is not null, # push it to queue with the # maximum value m if (node.left != None): q.append([ node.left, m ]) # If right child is not null, # push it to queue with the # maximum value m if (node.right != None): q.append([ node.right, m ]) # Returns the answer return count# Construct a Binary Treeroot = Noneroot = newNode(3)root.left = newNode(1)root.right = newNode(4)root.left.left = newNode(3)root.right.left = newNode(1)root.right.right = newNode(5)print(countNodes(root))# This code is contributed by rameshtravel07. |
C#
// C# program for the above approachusing System;using System.Collections;class GFG { // Node of the binary tree class Node { public int val; public Node left, right; public Node(int data) { val = data; left = right = null; } } // Function for creating new node static Node newNode(int data) { // Allocate memory for new node Node temp = new Node(data); // Return the created node return temp; } // Function to perform the DFS Traversal // to find the number of paths having // root to node X has value at most X static int countNodes(Node root) { // Initialize queue Queue q = new Queue(); int m = Int32.MinValue; // Push root in queue with the // maximum value m q.Enqueue(new Tuple<Node, int>(root, m)); // Stores the count of good nodes int count = 0; // Traverse all nodes while (q.Count > 0) { // Store the front node of // the queue Tuple<Node, int> temp = (Tuple<Node, int>)q.Peek(); q.Dequeue(); Node node = temp.Item1; int num = temp.Item2; // Check is current node is // greater than the maximum // value in path from root to // the current node if (node.val >= num) count++; // Update the maximum value m m = Math.Max(node.val, num); // If left child is not null, // push it to queue with the // maximum value m if (node.left != null) q.Enqueue(new Tuple<Node, int>(node.left, m)); // If right child is not null, // push it to queue with the // maximum value m if (node.right != null) q.Enqueue(new Tuple<Node, int>(node.right, m)); } // Returns the answer return count; } // Driver code static void Main() { // Construct a Binary Tree Node root = null; root = newNode(3); root.left = newNode(1); root.right = newNode(4); root.left.left = newNode(3); root.right.left = newNode(1); root.right.right = newNode(5); Console.Write(countNodes(root)); }}// This code is contributed by decode2207. |
Javascript
<script> // JavaScript program for the above approach class Node { constructor(data) { this.left = null; this.right = null; this.val = data; } } // Function for creating new node function newNode(data) { // Allocate memory for new node let temp = new Node(data); // Return the created node return temp; } // Function to perform the DFS Traversal // to find the number of paths having // root to node X has value at most X function countNodes(root) { // Initialize queue let q = []; let m = Number.MIN_VALUE; // Push root in queue with the // maximum value m q.push([ root, m ]); // Stores the count of good nodes let count = 0; // Traverse all nodes while (q.length > 0) { // Store the front node of // the queue let temp = q[0]; q.shift(); let node = temp[0]; let num = temp[1]; // Check is current node is // greater than the maximum // value in path from root to // the current node if (node.val >= num) count++; // Update the maximum value m m = Math.max(node.val, num); // If left child is not null, // push it to queue with the // maximum value m if (node.left) q.push([ node.left, m ]); // If right child is not null, // push it to queue with the // maximum value m if (node.right) q.push([ node.right, m ]); } // Returns the answer return count; } // Construct a Binary Tree let root = null; root = newNode(3); root.left = newNode(1); root.right = newNode(4); root.left.left = newNode(3); root.right.left = newNode(1); root.right.right = newNode(5); document.write(countNodes(root)); </script> |
Output:
4
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
