Given a Binary Tree, the task is to find the node from the given tree which has the maximum number of nodes in its subtree with values less than the value of that node. In the case of multiple possible nodes with the same number of maximum nodes, then return any such node.
Examples:
Input:
4
/ \
6 10
/ \ / \
2 3 7 14
/
5
Output: 6
Explanation:
Node with value 6 has the maximum of nodes which are less than 6 in the subtree of 6 as (2, 3, 5) i.e., 3.Input:
10
/
21
/ \
2 4
\
11Output: 21
Explanation:
Node with value 21 has the maximum of nodes which are less than 21 in the subtree of 21 as (2, 4, 11) i.e., 3.
Approach: The idea is to use the Post Order traversal. Below are the steps:
- Perform the Post Order Traversal on the given tree.
- Compare the nodes from the left sub-tree and the right sub-tree to its root value and if it is less than the root’s value and store the nodes which are less than the root node.
- Using the above step at each Node, find the number of nodes, then choose the node that has the maximum number of nodes whose keys are less than the current node.
- After the above traversal print that node having the maximum count of lesser node value than that nodes.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Stores the nodes to be deleted unordered_map< int , bool > mp; // Structure of a Tree node struct Node { int key; struct Node *left, *right; }; // 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 compare the current node // key with keys received from it left // & right tree by Post Order traversal vector< int > findNodes(Node* root, int & max_v, int & rootIndex) { // Base Case if (!root) { return vector< int >{}; } // Find nodes lesser than the current // root in the left subtree vector< int > left = findNodes(root->left, max_v, rootIndex); // Find nodes lesser than the current // root in the right subtree vector< int > right = findNodes(root->right, max_v, rootIndex); // Stores all the nodes less than // the current node's vector< int > combined; int count = 0; // Add the nodes which are less // than current node in left[] for ( int i = 0; i < left.size(); i++) { if (left[i] < root->key) { count += 1; } combined.push_back(left[i]); } // Add the nodes which are less // than current node in right[] for ( int i = 0; i < right.size(); i++) { if (right[i] < root->key) { count += 1; } combined.push_back(right[i]); } // Create a combined vector for // pass to it's parent combined.push_back(root->key); // Stores key that has maximum nodes if (count > max_v) { rootIndex = root->key; max_v = count; } // Return the vector of nodes return combined; } // Driver Code int main() { /* 3 / \ 4 6 / \ / \ 10 2 4 5 */ // Given Tree Node* root = newNode(3); root->left = newNode(4); root->right = newNode(6); root->right->left = newNode(4); root->right->right = newNode(5); root->left->left = newNode(10); root->left->right = newNode(2); int max_v = 0; int rootIndex = -1; // Function Call findNodes(root, max_v, rootIndex); // Print the node value cout << rootIndex; } |
Java
// Java program for // the above approach import java.util.*; class GFG{ // Stores the nodes to be deleted static HashMap<Integer, Boolean> mp = new HashMap<Integer, Boolean>(); static int max_v, rootIndex; // Structure of a Tree node static class Node { int key; Node left, right; }; // Function to create a new node static Node newNode( int key) { Node temp = new Node(); temp.key = key; temp.left = temp.right = null ; return (temp); } // Function to compare the current node // key with keys received from it left // & right tree by Post Order traversal static Vector<Integer> findNodes(Node root) { // Base Case if (root == null ) { return new Vector<Integer>(); } // Find nodes lesser than the current // root in the left subtree Vector<Integer> left = findNodes(root.left); // Find nodes lesser than the current // root in the right subtree Vector<Integer> right = findNodes(root.right); // Stores all the nodes less than // the current node's Vector<Integer> combined = new Vector<Integer>(); int count = 0 ; // Add the nodes which are less // than current node in left[] for ( int i = 0 ; i < left.size(); i++) { if (left.get(i) < root.key) { count += 1 ; } combined.add(left.get(i)); } // Add the nodes which are less // than current node in right[] for ( int i = 0 ; i < right.size(); i++) { if (right.get(i) < root.key) { count += 1 ; } combined.add(right.get(i)); } // Create a combined vector for // pass to it's parent combined.add(root.key); // Stores key that has maximum nodes if (count > max_v) { rootIndex = root.key; max_v = count; } // Return the vector of nodes return combined; } // Driver Code public static void main(String[] args) { /* 3 / \ 4 6 / \ / \ 10 2 4 5 */ // Given Tree Node root = newNode( 3 ); root.left = newNode( 4 ); root.right = newNode( 6 ); root.right.left = newNode( 4 ); root.right.right = newNode( 5 ); root.left.left = newNode( 10 ); root.left.right = newNode( 2 ); max_v = 0 ; rootIndex = - 1 ; // Function Call findNodes(root); // Print the node value System.out.print(rootIndex); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 program for the above approach # Stores the nodes to be deleted max_v = 0 rootIndex = 0 mp = {} # Structure of a Tree node class newNode: def __init__( self , key): self .key = key self .left = None self .right = None # Function to compare the current node # key with keys received from it left # & right tree by Post Order traversal def findNodes(root): global max_v global rootIndex global mp # Base Case if (root = = None ): return [] # Find nodes lesser than the current # root in the left subtree left = findNodes(root.left) # Find nodes lesser than the current # root in the right subtree right = findNodes(root.right) # Stores all the nodes less than # the current node's combined = [] count = 0 # Add the nodes which are less # than current node in left[] for i in range ( len (left)): if (left[i] < root.key): count + = 1 combined.append(left[i]) # Add the nodes which are less # than current node in right[] for i in range ( len (right)): if (right[i] < root.key): count + = 1 combined.append(right[i]) # Create a combined vector for # pass to it's parent combined.append(root.key) # Stores key that has maximum nodes if (count > max_v): rootIndex = root.key max_v = count # Return the vector of nodes return combined # Driver Code if __name__ = = '__main__' : ''' 3 / \ 4 6 / \ / \ 10 2 4 5 ''' # Given Tree root = None root = newNode( 3 ) root.left = newNode( 4 ) root.right = newNode( 6 ) root.right.left = newNode( 4 ) root.right.right = newNode( 5 ) root.left.left = newNode( 10 ) root.left.right = newNode( 2 ) max_v = 0 rootIndex = - 1 # Function Call findNodes(root) # Print the node value print (rootIndex) # This code is contributed by ipg2016107 |
C#
// C# program for // the above approach using System; using System.Collections.Generic; class GFG{ // Stores the nodes to be deleted static Dictionary< int , Boolean> mp = new Dictionary< int , Boolean>(); static int max_v, rootIndex; // Structure of a Tree node class Node { public int key; public Node left, right; }; // Function to create a new node static Node newNode( int key) { Node temp = new Node(); temp.key = key; temp.left = temp.right = null ; return (temp); } // Function to compare the current node // key with keys received from it left // & right tree by Post Order traversal static List< int > findNodes(Node root) { // Base Case if (root == null ) { return new List< int >(); } // Find nodes lesser than the current // root in the left subtree List< int > left = findNodes(root.left); // Find nodes lesser than the current // root in the right subtree List< int > right = findNodes(root.right); // Stores all the nodes less than // the current node's List< int > combined = new List< int >(); int count = 0; // Add the nodes which are less // than current node in []left for ( int i = 0; i < left.Count; i++) { if (left[i] < root.key) { count += 1; } combined.Add(left[i]); } // Add the nodes which are less // than current node in []right for ( int i = 0; i < right.Count; i++) { if (right[i] < root.key) { count += 1; } combined.Add(right[i]); } // Create a combined vector for // pass to it's parent combined.Add(root.key); // Stores key that has maximum nodes if (count > max_v) { rootIndex = root.key; max_v = count; } // Return the vector of nodes return combined; } // Driver Code public static void Main(String[] args) { /* 3 / \ 4 6 / \ / \ 10 2 4 5 */ // Given Tree Node root = newNode(3); root.left = newNode(4); root.right = newNode(6); root.right.left = newNode(4); root.right.right = newNode(5); root.left.left = newNode(10); root.left.right = newNode(2); max_v = 0; rootIndex = -1; // Function call findNodes(root); // Print the node value Console.Write(rootIndex); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // JavaScript program for the above approach // Stores the nodes to be deleted let mp = new Map(); let max_v, rootIndex; // Structure of a Tree node class Node { constructor(key) { this .left = null ; this .right = null ; this .key = key; } } // Function to create a new node function newNode(key) { let temp = new Node(key); return (temp); } // Function to compare the current node // key with keys received from it left // & right tree by Post Order traversal function findNodes(root) { // Base Case if (root == null ) { return []; } // Find nodes lesser than the current // root in the left subtree let left = findNodes(root.left); // Find nodes lesser than the current // root in the right subtree let right = findNodes(root.right); // Stores all the nodes less than // the current node's let combined = []; let count = 0; // Add the nodes which are less // than current node in left[] for (let i = 0; i < left.length; i++) { if (left[i] < root.key) { count += 1; } combined.push(left[i]); } // Add the nodes which are less // than current node in right[] for (let i = 0; i < right.length; i++) { if (right[i] < root.key) { count += 1; } combined.push(right[i]); } // Create a combined vector for // pass to it's parent combined.push(root.key); // Stores key that has maximum nodes if (count > max_v) { rootIndex = root.key; max_v = count; } // Return the vector of nodes return combined; } /* 3 / \ 4 6 / \ / \ 10 2 4 5 */ // Given Tree let root = newNode(3); root.left = newNode(4); root.right = newNode(6); root.right.left = newNode(4); root.right.right = newNode(5); root.left.left = newNode(10); root.left.right = newNode(2); max_v = 0; rootIndex = -1; // Function Call findNodes(root); // Print the node value document.write(rootIndex); </script> |
6
Time Complexity: O(N2)
Auxiliary Space: O(N)
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