Given a Binary Tree, the task is to count the number of nodes in the given Binary Tree such that the path from the root to that node contains node with value greater than or equal to that node.
Examples:
Input:
6
/ \
7 4
/ \ / \
3 7 1 2
Output: 5
Explanation:
Root node 6 is considered as its the only node
in the path from root to itself.
Node 4 has the minimum value in it's path 6->4.
Node 1 has the minimum value in it's path 6->4->1.
Node 2 has the minimum value in it's path 6->4->2.
Node 3 has the minimum value in it's path 6->7->3.
Input:
8
/ \
6 5
/ \ / \
6 7 3 9
Output: 5
Explanation:
Root node 8 is considered as its the only node
in the path from root to itself.
Node 6 has the minimum value in it's path 8->6.
Node 6 has the minimum value in it's path 8->6->6.
Node 5 has the minimum value in it's path 8->5.
Node 3 has the minimum value in it's path 8->5->3.
Approach: The idea is to do Preorder traversal on the given Binary Tree. Follow the steps below to solve the problem:
- Create a function to calculate the number of nodes that satisfy the given conditions.
- If the current node is NULL then return to the previous node.
- Use a variable minNodeVal to store the minimum node value along the path from the root to the current node.
- If the value of the current node is less than or equal to minNodeVal then increase the final count by 1 and update the value of minNodeVal.
- Call the function for the left and right child of the current node and repeat this process for every node to get the total count of the required nodes.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Structure of a tree nodestruct Node { int key; Node *left, *right;};// Function to create new tree nodeNode* newNode(int key){ Node* temp = new Node; temp->key = key; temp->left = temp->right = NULL; return temp;}// Function to find the total// number of required nodesvoid countReqNodes(Node* root, int minNodeVal, int& ans){ // If current node is null then // return to the parent node if (root == NULL) return; // Check if current node value is // less than or equal to minNodeVal if (root->key <= minNodeVal) { // Update the value of minNodeVal minNodeVal = root->key; // Update the count ans++; } // Go to the left subtree countReqNodes(root->left, minNodeVal, ans); // Go to the right subtree countReqNodes(root->right, minNodeVal, ans);}// Driver Codeint main(){ /* Binary Tree creation 8 / \ / \ 6 5 / \ / \ / \ / \ 6 7 3 9 */ Node* root = newNode(8); root->left = newNode(6); root->right = newNode(5); root->left->left = newNode(6); root->left->right = newNode(7); root->right->left = newNode(3); root->right->right = newNode(9); int ans = 0, minNodeVal = INT_MAX; // Function Call countReqNodes(root, minNodeVal, ans); // Print the result cout << ans; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Structure of a tree nodestatic class Node { int key; Node left, right;};// Function to create new tree nodestatic Node newNode(int key){ Node temp = new Node(); temp.key = key; temp.left = temp.right = null; return temp;}static int ans;// Function to find the total// number of required nodesstatic void countReqNodes(Node root, int minNodeVal){ // If current node is null then // return to the parent node if (root == null) return; // Check if current node value is // less than or equal to minNodeVal if (root.key <= minNodeVal) { // Update the value of minNodeVal minNodeVal = root.key; // Update the count ans++; } // Go to the left subtree countReqNodes(root.left, minNodeVal); // Go to the right subtree countReqNodes(root.right, minNodeVal);}// Driver Codepublic static void main(String[] args){ /* Binary Tree creation 8 / \ / \ 6 5 / \ / \ / \ / \ 6 7 3 9 */ Node root = newNode(8); root.left = newNode(6); root.right = newNode(5); root.left.left = newNode(6); root.left.right = newNode(7); root.right.left = newNode(3); root.right.right = newNode(9); int minNodeVal = Integer.MAX_VALUE; ans = 0; // Function Call countReqNodes(root, minNodeVal); // Print the result System.out.print(ans);}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approachimport sysans = 0# Class of a tree nodeclass Node: def __init__(self, key): self.key = key self.left = None self.right = None# Function to find the total# number of required nodesdef countReqNodes(root, minNodeVal): global ans # If current node is null then # return to the parent node if root == None: return # Check if current node value is # less than or equal to minNodeVal if root.key <= minNodeVal: # Update the value of minNodeVal minNodeVal = root.key # Update the count ans += 1 # Go to the left subtree countReqNodes(root.left, minNodeVal) # Go to the right subtree countReqNodes(root.right, minNodeVal)# Driver Codeif __name__ == '__main__': # Binary Tree creation # 8 # / \ # / \ # 6 5 # / \ / \ # / \ / \ # 6 7 3 9 # root = Node(8) root.left = Node(6) root.right = Node(5) root.left.left = Node(6) root.left.right = Node(7) root.right.left = Node(3) root.right.right = Node(9) minNodeVal = sys.maxsize # Function Call countReqNodes(root, minNodeVal) # Print the result print(ans)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Structure of a tree nodepublic class Node { public int key; public Node left, right;};// Function to create new tree nodestatic Node newNode(int key){ Node temp = new Node(); temp.key = key; temp.left = temp.right = null; return temp;}static int ans;// Function to find the total// number of required nodesstatic void countReqNodes(Node root, int minNodeVal){ // If current node is null then // return to the parent node if (root == null) return; // Check if current node value is // less than or equal to minNodeVal if (root.key <= minNodeVal) { // Update the value of minNodeVal minNodeVal = root.key; // Update the count ans++; } // Go to the left subtree countReqNodes(root.left, minNodeVal); // Go to the right subtree countReqNodes(root.right, minNodeVal);}// Driver Codepublic static void Main(String[] args){ /* Binary Tree creation 8 / \ / \ 6 5 / \ / \ / \ / \ 6 7 3 9 */ Node root = newNode(8); root.left = newNode(6); root.right = newNode(5); root.left.left = newNode(6); root.left.right = newNode(7); root.right.left = newNode(3); root.right.right = newNode(9); int minNodeVal = int.MaxValue; ans = 0; // Function Call countReqNodes(root, minNodeVal); // Print the result Console.Write(ans);}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for the above approach// Structure of tree nodeclass Node{ constructor(key) { this.left = null; this.right = null; this.key = key; }}// Function to create new tree nodefunction newNode(key){ let temp = new Node(key); return temp;}let ans;// Function to find the total// number of required nodesfunction countReqNodes(root, minNodeVal){ // If current node is null then // return to the parent node if (root == null) return; // Check if current node value is // less than or equal to minNodeVal if (root.key <= minNodeVal) { // Update the value of minNodeVal minNodeVal = root.key; // Update the count ans++; } // Go to the left subtree countReqNodes(root.left, minNodeVal); // Go to the right subtree countReqNodes(root.right, minNodeVal);}// Driver code/* Binary Tree creation 8 / \ / \ 6 5 / \ / \ / \ / \ 6 7 3 9*/let root = newNode(8);root.left = newNode(6);root.right = newNode(5);root.left.left = newNode(6);root.left.right = newNode(7);root.right.left = newNode(3);root.right.right = newNode(9);let minNodeVal = Number.MAX_VALUE;ans = 0; // Function CallcountReqNodes(root, minNodeVal); // Print the resultdocument.write(ans);// This code is contributed by suresh07</script> |
Output:
5
Time Complexity: O(N)
Auxiliary Space: O(1)
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