We have a simple binary tree and we have to print the top 3 largest elements present in the binary tree.
Examples:
Input :
1
/ \
2 3
/ \ / \
4 5 4 5
Output :Three largest elements are 5 4 3
Approach We can simply take three variables first, second, third to store the first largest, second largest, third largest respectively and perform preorder traversal and each time we will update the elements accordingly.
This approach will take O(n) time only.
Algorithm:
1- Take 3 variables first, second, third
2- Perform a preorder traversal
if (root==NULL)
return
if root's data is larger than first
update third with second
second with first
first with root's data
else if root's data is larger than
second and not equal to first
update third with second
second with root's data
else if root's data is larger than
third and not equal to first &
second
update third with root's data
3- call preorder for root->left
4- call preorder for root->right
Implementation:
C++
// CPP program to find largest three elements in// a binary tree.#include <bits/stdc++.h>using namespace std;struct Node { int data; struct Node *left; struct Node *right;};/* Helper function that allocates a new Node with the given data and NULL left and right pointers. */struct Node *newNode(int data) { struct Node *node = new Node; node->data = data; node->left = NULL; node->right = NULL; return (node);}// function to find three largest elementvoid threelargest(Node *root, int &first, int &second, int &third) { if (root == NULL) return; // if data is greater than first large number // update the top three list if (root->data > first) { third = second; second = first; first = root->data; } // if data is greater than second large number // and not equal to first update the bottom // two list else if (root->data > second && root->data != first) { third = second; second = root->data; } // if data is greater than third large number // and not equal to first & second update the // third highest list else if (root->data > third && root->data != first && root->data != second) third = root->data; threelargest(root->left, first, second, third); threelargest(root->right, first, second, third);}// driver functionint main() { struct Node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->left = newNode(4); root->right->right = newNode(5); int first = 0, second = 0, third = 0; threelargest(root, first, second, third); cout << "three largest elements are " << first << " " << second << " " << third; return 0;} |
Java
// Java program to find largest three elements // in a binary tree.import java.util.*;class GFG {static class Node { int data; Node left; Node right;};static int first, second, third;/* Helper function that allocates a new Node with the given data and null left and right pointers. */static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return (node);}// function to find three largest elementstatic void threelargest(Node root) {if (root == null) return;// if data is greater than first large number// update the top three listif (root.data > first) { third = second; second = first; first = root.data;}// if data is greater than second large number// and not equal to first update the bottom // two listelse if (root.data > second && root.data != first){ third = second; second = root.data;}// if data is greater than third large number// and not equal to first & second update the // third highest listelse if (root.data > third && root.data != first && root.data != second) third = root.data;threelargest(root.left);threelargest(root.right);}// driver functionpublic static void main(String[] args) { Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(4); root.right.right = newNode(5); first = 0; second = 0; third = 0; threelargest(root); System.out.print("three largest elements are " + first + " " + second + " " + third);}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to find largest three # elements in a binary tree. # Helper function that allocates a new # Node with the given data and None# left and right pointers. class newNode: def __init__(self, data): self.data = data self.left = None self.right = None # function to find three largest element def threelargest(root, first, second, third): if (root == None): return # if data is greater than first large # number update the top three list if (root.data > first[0]): third[0] = second[0] second[0] = first[0] first[0] = root.data # if data is greater than second large # number and not equal to first update # the bottom two list elif (root.data > second[0] and root.data != first[0]): third[0] = second[0] second[0] = root.data # if data is greater than third large # number and not equal to first & second # update the third highest list elif (root.data > third[0] and root.data != first[0] and root.data != second[0]): third[0] = root.data threelargest(root.left, first, second, third) threelargest(root.right, first, second, third)# Driver Codeif __name__ == '__main__': root = newNode(1) root.left = newNode(2) root.right = newNode(3) root.left.left = newNode(4) root.left.right = newNode(5) root.right.left = newNode(4) root.right.right = newNode(5) first = [0] second = [0] third = [0] threelargest(root, first, second, third) print("three largest elements are", first[0], second[0], third[0])# This code is contributed by PranchalK |
C#
// C# program to find largest three elements // in a binary tree.using System;class GFG {public class Node { public int data; public Node left; public Node right;};static int first, second, third;/* Helper function that allocates a new Node with the given data and null left and right pointers. */static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return (node);}// function to find three largest elementstatic void threelargest(Node root) { if (root == null) return; // if data is greater than first large number // update the top three list if (root.data > first) { third = second; second = first; first = root.data; } // if data is greater than second large number // and not equal to first update the bottom // two list else if (root.data > second && root.data != first) { third = second; second = root.data; } // if data is greater than third large number // and not equal to first & second update the // third highest list else if (root.data > third && root.data != first && root.data != second) third = root.data; threelargest(root.left); threelargest(root.right);}// Driver Codepublic static void Main(String[] args) { Node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(4); root.right.right = newNode(5); first = 0; second = 0; third = 0; threelargest(root); Console.WriteLine("three largest elements are " + first + " " + second + " " + third);}}// This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript program to find largest // three elements in a binary tree. class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } let first, second, third; /* Helper function that allocates a new Node with the given data and null left and right pointers. */ function newNode(data) { let node = new Node(data); return (node); } // function to find three largest element function threelargest(root) { if (root == null) return; // if data is greater than first large number // update the top three list if (root.data > first) { third = second; second = first; first = root.data; } // if data is greater than second large number // and not equal to first update the bottom // two list else if (root.data > second && root.data != first) { third = second; second = root.data; } // if data is greater than third large number // and not equal to first & second update the // third highest list else if (root.data > third && root.data != first && root.data != second) third = root.data; threelargest(root.left); threelargest(root.right); } let root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.left = newNode(4); root.right.right = newNode(5); first = 0; second = 0; third = 0; threelargest(root); document.write("three largest elements are " + first + " " + second + " " + third); </script> |
Output:
three largest elements are 5 4 3
Time Complexity: O(N)
Auxiliary Space: O(1)
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