Find the maximum node at a given level in a binary tree
Given a Binary Tree and a Level. The task is to find the node with the maximum value at that given level.
The idea is to traverse the tree along depth recursively and return the nodes once the required level is reached and then return the maximum of left and right subtrees for each subsequent call. So that the last call will return the node with maximum value among all nodes at the given level.
Below is the step by step algorithm:
- Perform DFS traversal and every time decrease the value of level by 1 and keep traversing to the left and right subtrees recursively.
- When value of level becomes 0, it means we are on the given level, then return root->data.
- Find the maximum between the two values returned by left and right subtrees and return the maximum.
Below is the implementation of above approach:
C++
// CPP program to find the node with // maximum value at a given level #include <iostream> using namespace std; // Tree node struct Node { int data; struct Node *left, *right; }; // Utility function to create a new Node struct Node* newNode(int val) { struct Node* temp = new Node; temp->left = NULL; temp->right = NULL; temp->data = val; return temp; } // function to find the maximum value // at given level int maxAtLevel(struct Node* root, int level) { // If the tree is empty if (root == NULL) return 0; // if level becomes 0, it means we are on // any node at the given level if (level == 0) return root->data; int x = maxAtLevel(root->left, level - 1); int y = maxAtLevel(root->right, level - 1); // return maximum of two return max(x, y); } // Driver code int main() { // Creating the tree struct Node* root = NULL; root = newNode(45); root->left = newNode(46); root->left->left = newNode(18); root->left->left->left = newNode(16); root->left->left->right = newNode(23); root->left->right = newNode(17); root->left->right->left = newNode(24); root->left->right->right = newNode(21); root->right = newNode(15); root->right->left = newNode(22); root->right->left->left = newNode(37); root->right->left->right = newNode(41); root->right->right = newNode(19); root->right->right->left = newNode(49); root->right->right->right = newNode(29); int level = 3; cout << maxAtLevel(root, level); return 0; } |
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Java
// Java program to find the // node with maximum value // at a given level import java.util.*; class GFG { // Tree node static class Node { int data; Node left, right; } // Utility function to // create a new Node static Node newNode(int val) { Node temp = new Node(); temp.left = null; temp.right = null; temp.data = val; return temp; } // function to find // the maximum value // at given level static int maxAtLevel(Node root, int level) { // If the tree is empty if (root == null) return 0; // if level becomes 0, // it means we are on // any node at the given level if (level == 0) return root.data; int x = maxAtLevel(root.left, level - 1); int y = maxAtLevel(root.right, level - 1); // return maximum of two return Math.max(x, y); } // Driver code public static void main(String args[]) { // Creating the tree Node root = null; root = newNode(45); root.left = newNode(46); root.left.left = newNode(18); root.left.left.left = newNode(16); root.left.left.right = newNode(23); root.left.right = newNode(17); root.left.right.left = newNode(24); root.left.right.right = newNode(21); root.right = newNode(15); root.right.left = newNode(22); root.right.left.left = newNode(37); root.right.left.right = newNode(41); root.right.right = newNode(19); root.right.right.left = newNode(49); root.right.right.right = newNode(29); int level = 3; System.out.println(maxAtLevel(root, level)); } } // This code is contributed // by Arnab Kundu |
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Python3
# Python3 program to find the node # with maximum value at a given level # Helper function that allocates a new # node with the given data and None # left and right poers. class newNode: # Constructor to create a new node def __init__(self, data): self.data = data self.left = None self.right = None # function to find the maximum # value at given level def maxAtLevel(root, level): # If the tree is empty if (root == None) : return 0 # if level becomes 0, it means we # are on any node at the given level if (level == 0) : return root.data x = maxAtLevel(root.left, level - 1) y = maxAtLevel(root.right, level - 1) # return maximum of two return max(x, y) # Driver Code if __name__ == '__main__': """ Let us create Binary Tree shown in above example """ root = newNode(45) root.left = newNode(46) root.left.left = newNode(18) root.left.left.left = newNode(16) root.left.left.right = newNode(23) root.left.right = newNode(17) root.left.right.left = newNode(24) root.left.right.right = newNode(21) root.right = newNode(15) root.right.left = newNode(22) root.right.left.left = newNode(37) root.right.left.right = newNode(41) root.right.right = newNode(19) root.right.right.left = newNode(49) root.right.right.right = newNode(29) level = 3 print(maxAtLevel(root, level)) # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
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C#
// C# program to find the // node with maximum value // at a given level using System; class GFG { // Tree node class Node { public int data; public Node left, right; } // Utility function to // create a new Node static Node newNode(int val) { Node temp = new Node(); temp.left = null; temp.right = null; temp.data = val; return temp; } // function to find // the maximum value // at given level static int maxAtLevel(Node root, int level) { // If the tree is empty if (root == null) return 0; // if level becomes 0, // it means we are on // any node at the given level if (level == 0) return root.data; int x = maxAtLevel(root.left, level - 1); int y = maxAtLevel(root.right, level - 1); // return maximum of two return Math.Max(x, y); } // Driver code public static void Main(String []args) { // Creating the tree Node root = null; root = newNode(45); root.left = newNode(46); root.left.left = newNode(18); root.left.left.left = newNode(16); root.left.left.right = newNode(23); root.left.right = newNode(17); root.left.right.left = newNode(24); root.left.right.right = newNode(21); root.right = newNode(15); root.right.left = newNode(22); root.right.left.left = newNode(37); root.right.left.right = newNode(41); root.right.right = newNode(19); root.right.right.left = newNode(49); root.right.right.right = newNode(29); int level = 3; Console.WriteLine(maxAtLevel(root, level)); } } // This code is contributed by 29AjayKumar |
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Output:
49
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