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Create a wave array from the given Binary Search Tree

  • Difficulty Level : Easy
  • Last Updated : 29 Nov, 2021

Given a Binary Search Tree, the task is to create a wave array from the given Binary Search Tree. An array arr[0..n-1] is called a wave array if arr[0] >= arr[1] <= arr[2] >= arr[3] <= arr[4] >= …

Examples:

Input:

Output: 4 2 8 6 12 10 14
Explanation: The above mentioned array {4, 2, 8, 6, 12, 10, 14} is one of the many valid wave arrays.

Input:

Output: 4 2 8 6 12 

Approach: The given problem can be solved by the observation that the Inorder Traversal of the Binary Search Tree gives nodes in non-decreasing order. Therefore, store the inorder traversal of the given tree into a vector. Since the vector contains elements in sorted order, it can be converted into a wave array by swapping the adjacent elements for all elements in the range [0, N) using the approach discussed in this article.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Node of the Binary Search tree
struct Node {
    int data;
    Node* right;
    Node* left;
 
    // Constructor
    Node(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
 
// Function to convert Binary Search
// Tree into a wave Array
void toWaveArray(Node* root)
{
    // Stores the final wave array
    vector<int> waveArr;
 
    stack<Node*> s;
    Node* curr = root;
 
    // Perform the Inorder traversal
    // of the given BST
    while (curr != NULL || s.empty() == false) {
 
        // Reach the left most Node of
        // the curr Node
        while (curr != NULL) {
 
            // Place pointer to a tree node
            // in stack before traversing
            // the node's left subtree
            s.push(curr);
            curr = curr->left;
        }
        curr = s.top();
        s.pop();
 
        // Insert into wave array
        waveArr.push_back(curr->data);
 
        // Visit the right subtree
        curr = curr->right;
    }
 
    // Convert sorted array into wave array
    for (int i = 0;
         i + 1 < waveArr.size(); i += 2) {
        swap(waveArr[i], waveArr[i + 1]);
    }
 
    // Print the answer
    for (int i = 0; i < waveArr.size(); i++) {
        cout << waveArr[i] << " ";
    }
}
 
// Driver Code
int main()
{
    Node* root = new Node(8);
    root->left = new Node(4);
    root->right = new Node(12);
    root->right->left = new Node(10);
    root->right->right = new Node(14);
    root->left->left = new Node(2);
    root->left->right = new Node(6);
 
    toWaveArray(root);
 
    return 0;
}


Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Node of the Binary Search tree
static class Node {
    int data;
    Node right;
    Node left;
 
    // Constructor
    Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
// Function to convert Binary Search
// Tree into a wave Array
static void toWaveArray(Node root)
{
   
    // Stores the final wave array
    Vector<Integer> waveArr = new Vector<>();
 
    Stack<Node> s = new Stack<>();
    Node curr = root;
 
    // Perform the Inorder traversal
    // of the given BST
    while (curr != null || s.isEmpty() == false) {
 
        // Reach the left most Node of
        // the curr Node
        while (curr != null) {
 
            // Place pointer to a tree node
            // in stack before traversing
            // the node's left subtree
            s.add(curr);
            curr = curr.left;
        }
        curr = s.peek();
        s.pop();
 
        // Insert into wave array
        waveArr.add(curr.data);
 
        // Visit the right subtree
        curr = curr.right;
    }
 
    // Convert sorted array into wave array
    for (int i = 0;   i + 1 < waveArr.size(); i += 2) {
        int t = waveArr.get(i);
        waveArr.set(i, waveArr.get(i+1));
        waveArr.set(i+1, t);
         
    }
 
    // Print the answer
    for (int i = 0; i < waveArr.size(); i++) {
        System.out.print(waveArr.get(i)+ " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    Node root = new Node(8);
    root.left = new Node(4);
    root.right = new Node(12);
    root.right.left = new Node(10);
    root.right.right = new Node(14);
    root.left.left = new Node(2);
    root.left.right = new Node(6);
 
    toWaveArray(root);
 
}
}
 
// This code is contributed by umadevi9616


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
// Node of the Binary Search tree
public class Node {
    public int data;
    public Node right;
    public Node left;
 
    // Constructor
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
// Function to convert Binary Search
// Tree into a wave Array
static void toWaveArray(Node root)
{
   
    // Stores the readonly wave array
    List<int> waveArr = new List<int>();
 
    Stack<Node> s = new Stack<Node>();
    Node curr = root;
 
    // Perform the Inorder traversal
    // of the given BST
    while (curr != null || s.Count!=0 ) {
 
        // Reach the left most Node of
        // the curr Node
        while (curr != null) {
 
            // Place pointer to a tree node
            // in stack before traversing
            // the node's left subtree
            s.Push(curr);
            curr = curr.left;
        }
        curr = s.Peek();
        s.Pop();
 
        // Insert into wave array
        waveArr.Add(curr.data);
 
        // Visit the right subtree
        curr = curr.right;
    }
 
    // Convert sorted array into wave array
    for (int i = 0;   i + 1 < waveArr.Count; i += 2) {
        int t = waveArr[i];
        waveArr[i]= waveArr[i+1];
        waveArr[i+1]= t;
         
    }
 
    // Print the answer
    for (int i = 0; i < waveArr.Count; i++) {
        Console.Write(waveArr[i]+ " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    Node root = new Node(8);
    root.left = new Node(4);
    root.right = new Node(12);
    root.right.left = new Node(10);
    root.right.right = new Node(14);
    root.left.left = new Node(2);
    root.left.right = new Node(6);
 
    toWaveArray(root);
}
}
 
// This code is contributed by umadevi9616


Javascript




<script>
        // JavaScript Program to implement
        // the above approach
        class Node {
            constructor(data) {
                this.data = data;
                this.left = this.right = null;
            }
        }
        // Function to convert Binary Search
        // Tree into a wave Array
        function toWaveArray(root) {
            // Stores the final wave array
            let waveArr = [];
 
            let s = [];
            let curr = root;
 
            // Perform the Inorder traversal
            // of the given BST
            while (curr != null || s.length != 0) {
 
                // Reach the left most Node of
                // the curr Node
                while (curr != null) {
 
                    // Place pointer to a tree node
                    // in stack before traversing
                    // the node's left subtree
                    s.push(curr);
                    curr = curr.left;
                }
                curr = s[s.length - 1];
                s.pop();
 
                // Insert into wave array
                waveArr.push(curr.data);
 
                // Visit the right subtree
                curr = curr.right;
            }
 
            // Convert sorted array into wave array
            for (let i = 0;
                i + 1 < waveArr.length; i += 2) {
                let temp = waveArr[i]
                waveArr[i] = waveArr[i + 1]
                waveArr[i + 1] = temp
            }
 
            // Print the answer
            for (let i = 0; i < waveArr.length; i++) {
                document.write(waveArr[i] + " ");
            }
        }
 
        // Driver Code
        let root = new Node(8);
        root.left = new Node(4);
        root.right = new Node(12);
        root.right.left = new Node(10);
        root.right.right = new Node(14);
        root.left.left = new Node(2);
        root.left.right = new Node(6);
 
        toWaveArray(root);
 
     // This code is contributed by Potta Lokesh
 
    </script>


Output: 

4 2 8 6 12 10 14

 

Time Complexity: O(N)
Auxiliary Space: O(N)


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