Count of nodes that are greater than Ancestors
Given the root of a tree, the task is to find the count of nodes which are greater than all of its ancestors.
Examples:
Input: 4 / \ 5 2 / \ 3 6 Output: 3 The nodes are 4, 5 and 6. Input: 10 / \ 8 6 \ \ 3 5 / 1 Output: 1
Approach: The problem can be solved using dfs. In every function call, pass a variable maxx which stores the maximum among all the nodes traversed so far, and every node whose value is greater than maxx is the node that satisfies the given condition. Hence, increment the count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Structure for the node of the tree struct Tree { int val; Tree* left; Tree* right; Tree( int _val) { val = _val; left = NULL; right = NULL; } }; // Dfs Function void dfs(Tree* node, int maxx, int & count) { // Base case if (node == NULL) { return ; } else { // Increment the count if the current // node's value is greater than the // maximum value in it's ancestors if (node->val > maxx) count++; // Left traversal dfs(node->left, max(maxx, node->val), count); // Right traversal dfs(node->right, max(maxx, node->val), count); } } // Driver code int main() { Tree* root = new Tree(4); root->left = new Tree(5); root->right = new Tree(2); root->right->left = new Tree(3); root->right->right = new Tree(6); // To store the required count int count = 0; dfs(root, INT_MIN, count); cout << count; return 0; } |
Java
// Java implementation of the approach class GFG { static int count; // Structure for the node of the tree static class Tree { int val; Tree left; Tree right; Tree( int _val) { val = _val; left = null ; right = null ; } }; // Dfs Function static void dfs(Tree node, int maxx) { // Base case if (node == null ) { return ; } else { // Increment the count if the current // node's value is greater than the // maximum value in it's ancestors if (node.val > maxx) count++; // Left traversal dfs(node.left, Math.max(maxx, node.val)); // Right traversal dfs(node.right, Math.max(maxx, node.val)); } } // Driver code public static void main(String[] args) { Tree root = new Tree( 4 ); root.left = new Tree( 5 ); root.right = new Tree( 2 ); root.right.left = new Tree( 3 ); root.right.right = new Tree( 6 ); // To store the required count count = 0 ; dfs(root, Integer.MIN_VALUE); System.out.print(count); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program for the # above approach from collections import deque # A Tree node class Tree: def __init__( self , x): self .val = x self .left = None self .right = None count = 0 # Dfs Function def dfs(node, maxx): global count # Base case if (node = = None ): return else : # Increment the count if # the current node's value # is greater than the maximum # value in it's ancestors if (node.val > maxx): count + = 1 # Left traversal dfs(node.left, max (maxx, node.val)) # Right traversal dfs(node.right, max (maxx, node.val)) # Driver code if __name__ = = '__main__' : root = Tree( 4 ) root.left = Tree( 5 ) root.right = Tree( 2 ) root.right.left = Tree( 3 ) root.right.right = Tree( 6 ) # To store the required # count count = 0 dfs(root, - 10 * * 9 ) print (count) # This code is contributed by Mohit Kumar 29 |
C#
// C# implementation of the approach using System; class GFG { static int count; // Structure for the node of the tree public class Tree { public int val; public Tree left; public Tree right; public Tree( int _val) { val = _val; left = null ; right = null ; } }; // Dfs Function static void dfs(Tree node, int maxx) { // Base case if (node == null ) { return ; } else { // Increment the count if the current // node's value is greater than the // maximum value in it's ancestors if (node.val > maxx) count++; // Left traversal dfs(node.left, Math.Max(maxx, node.val)); // Right traversal dfs(node.right, Math.Max(maxx, node.val)); } } // Driver code public static void Main(String[] args) { Tree root = new Tree(4); root.left = new Tree(5); root.right = new Tree(2); root.right.left = new Tree(3); root.right.right = new Tree(6); // To store the required count count = 0; dfs(root, int .MinValue); Console.Write(count); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript implementation of the approach let count=0; // Structure for the node of the tree class Tree { constructor(val) { this .val=val; this .left= this .right= null ; } } // Dfs Function function dfs(node,maxx) { // Base case if (node == null ) { return ; } else { // Increment the count if the current // node's value is greater than the // maximum value in it's ancestors if (node.val > maxx) count++; // Left traversal dfs(node.left, Math.max(maxx, node.val)); // Right traversal dfs(node.right, Math.max(maxx, node.val)); } } // Driver code let root = new Tree(4); root.left = new Tree(5); root.right = new Tree(2); root.right.left = new Tree(3); root.right.right = new Tree(6); // To store the required count count = 0; dfs(root, Number.MIN_VALUE); document.write(count); // This code is contributed by unknown2108 </script> |
Output
3
Time Complexity: O(N) where N is the number of nodes in given binary tree.
Space Complexity: O(h) where h is the height of binary tree due to stack recursion call.
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