Optimal sequence for AVL tree insertion (without any rotations)

Given an array of integers, the task is to find the sequence in which these integers should be added to an AVL tree such that no rotations are required to balance the tree.

Examples :

Input : array = {1, 2, 3}
Output : 2 1 3

Input : array = {2, 4, 1, 3, 5, 6, 7}
Output : 4 2 6 1 3 5 7


Approach :

  • Sort the given array of integers.
  • Create the AVL tree from the sorted array by following the approach described here.
  • Now, find the level order traversal of the tree which is the required sequence.
  • Adding numbers in the sequence found in the previous step will always maintain the height balance property of all the nodes in the tree.

Below is the implementation of the above approach :

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
/* A Binary Tree node */
struct TNode {
    int data;
    struct TNode* left;
    struct TNode* right;
};
  
struct TNode* newNode(int data);
  
/* Function to construct AVL tree 
   from a sorted array */
struct TNode* sortedArrayToBST(int arr[], int start, int end)
{
    /* Base Case */
    if (start > end)
        return NULL;
  
    /* Get the middle element 
       and make it root */
    int mid = (start + end) / 2;
    struct TNode* root = newNode(arr[mid]);
  
    /* Recursively construct the 
       left subtree and make it 
       left child of root */
    root->left = sortedArrayToBST(arr, start, mid - 1);
  
    /* Recursively construct the 
       right subtree and make it 
       right child of root */
    root->right = sortedArrayToBST(arr, mid + 1, end);
  
    return root;
}
  
/* Helper function that allocates
   a new node with the given data 
   and NULL to the left and 
   the right pointers. */
struct TNode* newNode(int data)
{
    struct TNode* node = (struct TNode*)
        malloc(sizeof(struct TNode));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
  
    return node;
}
  
// This function is used for testing purpose
void printLevelOrder(TNode *root) 
    if (root == NULL)  return
  
    queue<TNode *> q; 
    q.push(root); 
    
    while (q.empty() == false
    
        TNode *node = q.front(); 
        cout << node->data << " "
        q.pop(); 
        if (node->left != NULL) 
            q.push(node->left); 
        if (node->right != NULL) 
            q.push(node->right); 
    
}   
  
/* Driver program to 
test above functions */
int main()
{
  
    // Assuming the array is sorted
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    /* Convert List to AVL tree */
    struct TNode* root = sortedArrayToBST(arr, 0, n - 1);
    printLevelOrder(root);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

    
// Java implementation of the approach
import java.util.*;
class solution
{
  
/* A Binary Tree node */
 static  class TNode {
    int data;
     TNode left;
     TNode right;
}
  
  
/* Function to con AVL tree 
   from a sorted array */
 static TNode sortedArrayToBST(int arr[], int start, int end)
{
    /* Base Case */
    if (start > end)
        return null;
  
    /* Get the middle element 
       and make it root */
    int mid = (start + end) / 2;
     TNode root = newNode(arr[mid]);
  
    /* Recursively con the 
       left subtree and make it 
       left child of root */
    root.left = sortedArrayToBST(arr, start, mid - 1);
  
    /* Recursively con the 
       right subtree and make it 
       right child of root */
    root.right = sortedArrayToBST(arr, mid + 1, end);
  
    return root;
}
  
/* Helper function that allocates
   a new node with the given data 
   and null to the left and 
   the right pointers. */
static  TNode newNode(int data)
{
     TNode node = new TNode();
    node.data = data;
    node.left = null;
    node.right = null;
  
    return node;
}
  
// This function is used for testing purpose
static void printLevelOrder(TNode root) 
    if (root == nullreturn
  
    Queue<TNode > q= new LinkedList<TNode>(); 
    q.add(root); 
    
    while (q.size()>0
    
        TNode node = q.element(); 
        System.out.print( node.data + " "); 
        q.remove(); 
        if (node.left != null
            q.add(node.left); 
        if (node.right != null
            q.add(node.right); 
    
}   
  
/* Driver program to 
test above functions */
public static void main(String args[])
{
  
    // Assuming the array is sorted
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.length;
  
    /* Convert List to AVL tree */
     TNode root = sortedArrayToBST(arr, 0, n - 1);
    printLevelOrder(root);
  
}
}
//contributed by Arnab Kundu

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 code to print order of 
# insertion into AVL tree to 
# ensure no rotations
  
# Tree Node
class Node: 
    def __init__(self, d): 
        self.data =
        self.left = None
        self.right = None
  
# Function to convert sorted array 
# to a balanced AVL Tree/BST
# Input : sorted array of integers 
# Output: root node of balanced AVL Tree/BST 
def sortedArrayToBST(arr): 
      
    if not arr: 
        return None
  
    # Find middle and get its floor value
    mid = int((len(arr)) / 2)
    root = Node(arr[mid]) 
      
    # Recursively construct the left 
    # and right subtree
    root.left = sortedArrayToBST(arr[:mid]) 
    root.right = sortedArrayToBST(arr[mid + 1:]) 
  
    # Return the root of the 
    # constructed tree
    return root 
  
# A utility function to print the
# Level Order Traversal of AVL Tree
# using a Queue
def printLevelOrder(root): 
    if not root: 
        return
      
    q = []
    q.append(root)
  
    # Keep printing the top element and
    # adding to queue while it is not empty
    while q != []:
        node = q.pop(0)
        print(node.data, end=" ")
  
        # If left node exists, enqueue it
        if node.left:
            q.append(node.left)
  
        # If right node exists, enqueue it 
        if node.right:
            q.append(node.right)
  
# Driver Code
arr = [1, 2, 3, 4, 5, 6, 7
root = sortedArrayToBST(arr) 
printLevelOrder(root) 
  
# This code is contributed 
# by Adikar Bharath 

chevron_right


Output:

4 2 6 1 3 5 7


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : andrew1234, lkcbharath