Given an array of N elements which denotes the array representation of binary heap, the task is to find the leaf nodes of this binary heap.
Examples:
Input:
arr[] = {1, 2, 3, 4, 5, 6, 7}
Output: 4 5 6 7
Explanation:
1
/ \
2 3
/ \ / \
4 5 6 7
Leaf nodes of the Binary Heap are:
4 5 6 7
Input:
arr[] = {1, 2, 3, 4, 5,
6, 7, 8, 9, 10}
Output: 6 7 8 9 10
Explanation:
1
/ \
2 3
/ \ / \
4 5 6 7
/ \ /
8 9 10
Leaf Nodes of the Binary Heap are:
6 7 8 9 10
Approach: The key observation in the problem is that the every leaf node of the Binary Heap will be at the Height H or H -1, If H is the height of the Binary Heap. Therefore, the leaf nodes can be computed as follows:
- Calculate the total height of the binary heap.
- Traverse the array in reverse order and compare the height of each node to the compute height H of the Binary Heap.
- If the height of the current node is H, then add the current node to the leaf nodes.
- Otherwise, If the height of current node is H-1 and there are no child nodes, then also add the node as leaf node.
Below is the implementation of the above approach:
C++
// C++ implementation to print// the leaf nodes of a Binary Heap#include <bits/stdc++.h>using namespace std;// Function to find the height of// complete binary treeint height(int N) { return (int)floor(log2(N + 1)); }// Function to pr the leaf nodesvoid prLeafNodes(vector<int> arrlist){ for (int i = arrlist.size() - 1; i >= -0; i--) cout << arrlist[i] << " ";}// Function to find the leaf// nodes of binary heapvoid findLeafNodes(int arr[], int n){ // Calculate the height of // the complete binary tree int h = height(n); vector<int> arrlist; for (int i = n - 1; i >= 0; i--) { if (height(i + 1) == h) arrlist.push_back(arr[i]); else if (height(i + 1) == h - 1 && n <= ((2 * i) + 1)) // if the height if h-1, // then there should not // be any child nodes arrlist.push_back(arr[i]); else break; } prLeafNodes(arrlist);}// Driver Codeint main(){ int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; int n = sizeof(arr) / sizeof(arr[0]); findLeafNodes(arr, n); return 0;}// This code is contributed by adityamaharshi21 |
Java
// Java implementation to print// the leaf nodes of a Binary Heapimport java.lang.*;import java.util.*;class GFG { // Function to calculate height // of the Binary heap with given // the count of the nodes static int height(int N) { return (int)Math.ceil( Math.log(N + 1) / Math.log(2)) - 1; } // Function to find the leaf // nodes of binary heap static void findLeafNodes( int arr[], int n) { // Calculate the height of // the complete binary tree int h = height(n); ArrayList<Integer> arrlist = new ArrayList<>(); for (int i = n - 1; i >= 0; i--) { if (height(i + 1) == h) { arrlist.add(arr[i]); } else if (height(i + 1) == h - 1 && n <= ((2 * i) + 1)) { // if the height if h-1, // then there should not // be any child nodes arrlist.add(arr[i]); } else { break; } } printLeafNodes(arrlist); } // Function to print the leaf nodes static void printLeafNodes( ArrayList<Integer> arrlist) { for (int i = arrlist.size() - 1; i >= 0; i--) { System.out.print( arrlist.get(i) + " "); } } // Driver Code public static void main(String[] args) { int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; findLeafNodes(arr, arr.length); }} |
C#
// C# implementation to print// the leaf nodes of a Binary Heapusing System;using System.Collections.Generic;class GFG{// Function to calculate height// of the Binary heap with given// the count of the nodesstatic int height(int N){ return (int)Math.Ceiling( Math.Log(N + 1) / Math.Log(2)) - 1;}// Function to find the leaf// nodes of binary heapstatic void findLeafNodes(int []arr, int n){ // Calculate the height of // the complete binary tree int h = height(n); List<int> arrlist = new List<int>(); for (int i = n - 1; i >= 0; i--) { if (height(i + 1) == h) { arrlist.Add(arr[i]); } else if (height(i + 1) == h - 1 && n <= ((2 * i) + 1)) { // if the height if h-1, // then there should not // be any child nodes arrlist.Add(arr[i]); } else { break; } } printLeafNodes(arrlist);}// Function to print the leaf nodesstatic void printLeafNodes(List<int> arrlist){ for (int i = arrlist.Count - 1; i >= 0; i--) { Console.Write(arrlist[i] + " "); }}// Driver Codepublic static void Main(String[] args){ int []arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }; findLeafNodes(arr, arr.Length);}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation to print# the leaf nodes of a Binary Heapimport mathdef height(N): return math.log(N + 1) // math.log(2) # Function to find the leaf# nodes of binary heapdef findLeafNodes(arr, n): # Calculate the height of # the complete binary tree h = height(n) arrlist = [] for i in range(n - 1,-1,-1): if (height(i + 1) == h): arrlist.append(arr[i]) elif (height(i + 1) == h - 1 and n <= ((2 * i) + 1)): # if the height if h-1, # then there should not # be any child nodes arrlist.append(arr[i]) else: break prLeafNodes(arrlist) # Function to pr the leaf nodesdef prLeafNodes(arrlist): for i in range(len(arrlist) - 1,-1,-1): print(arrlist[i],end=" ") # Driver Codearr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]findLeafNodes(arr, len(arr))# This code is contributed by shinjanpatra |
Javascript
<script>// JavaScript implementation to print// the leaf nodes of a Binary Heapfunction height(N){ return Math.floor(Math.log(N + 1) / Math.log(2))} // Function to find the leaf// nodes of binary heapfunction findLeafNodes(arr, n){ // Calculate the height of // the complete binary tree let h = height(n) let arrlist = [] for(let i = n - 1;i >= 0 ;i--){ if (height(i + 1) == h) arrlist.push(arr[i]) else if (height(i + 1) == h - 1 && n <= ((2 * i) + 1)) // if the height if h-1, // then there should not // be any child nodes arrlist.push(arr[i]) else break } prLeafNodes(arrlist)} // Function to pr the leaf nodesfunction prLeafNodes(arrlist){ for(let i = arrlist.length - 1 ; i>=-0; i--) document.write(arrlist[i]," ")} // Driver Codelet arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]findLeafNodes(arr, arr.length)// This code is contributed by shinjanpatra</script> |
Output:
6 7 8 9 10
Performance Analysis:
- Time Complexity: O(L), where L is the number of leaf nodes.
- Auxiliary Space: O(1)
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