Construct the BST (Binary Search Tree) from its given level order traversal.
Examples:
Input : arr[] = {7, 4, 12, 3, 6, 8, 1, 5, 10}
Output : BST:
7
/ \
4 12
/ \ /
3 6 8
/ / \
1 5 10
The idea is to use the Recursion:-
We know that the first element will always be the root of tree and second element will be the left child and third element will be the right child (if fall in the range), and so on for all the remaining elements.
1) First pick the first element of the array and make it root.
2) Pick the second element, if it’s value is smaller than root node value make it left child,
3) Else make it right child
4) Now recursively call the step (2) and step (3) to make a BST from its level Order Traversal.
Below is the implementation of above approach:
// C++ implementation to construct a BST // from its level order traversal #include <bits/stdc++.h> using namespace std;
// node of a BST struct Node
{ int data;
Node *left, *right;
}; // function to get a new node Node* getNode(int data)
{ // Allocate memory
Node *newNode =
(Node*)malloc(sizeof(Node));
// put in the data
newNode->data = data;
newNode->left = newNode->right = NULL;
return newNode;
} // function to construct a BST from // its level order traversal Node *LevelOrder(Node *root , int data)
{ if(root==NULL){
root = getNode(data);
return root;
}
if(data <= root->data)
root->left = LevelOrder(root->left, data);
else
root->right = LevelOrder(root->right, data);
return root;
} Node* constructBst(int arr[], int n)
{ if(n==0)return NULL;
Node *root =NULL;
for(int i=0;i<n;i++)
root = LevelOrder(root , arr[i]);
return root;
} // function to print the inorder traversal void inorderTraversal(Node* root)
{ if (!root)
return;
inorderTraversal(root->left);
cout << root->data << " ";
inorderTraversal(root->right);
} // Driver program to test above int main()
{ int arr[] = {7, 4, 12, 3, 6, 8, 1, 5, 10};
int n = sizeof(arr) / sizeof(arr[0]);
Node *root = constructBst(arr, n);
cout << "Inorder Traversal: ";
inorderTraversal(root);
return 0;
} |
// Java implementation to construct a BST // from its level order traversal class GFG
{ // node of a BST static class Node
{ int data;
Node left, right;
}; // function to get a new node static Node getNode(int data)
{ // Allocate memory
Node newNode = new Node();
// put in the data
newNode.data = data;
newNode.left = newNode.right = null;
return newNode;
} // function to construct a BST from // its level order traversal static Node LevelOrder(Node root , int data)
{ if(root == null)
{
root = getNode(data);
return root;
}
if(data <= root.data)
root.left = LevelOrder(root.left, data);
else
root.right = LevelOrder(root.right, data);
return root;
} static Node constructBst(int arr[], int n)
{ if(n == 0)return null;
Node root = null;
for(int i = 0; i < n; i++)
root = LevelOrder(root , arr[i]);
return root;
} // function to print the inorder traversal static void inorderTraversal(Node root)
{ if (root == null)
return;
inorderTraversal(root.left);
System.out.print( root.data + " ");
inorderTraversal(root.right);
} // Driver code public static void main(String args[])
{ int arr[] = {7, 4, 12, 3, 6, 8, 1, 5, 10};
int n = arr.length;
Node root = constructBst(arr, n);
System.out.print( "Inorder Traversal: ");
inorderTraversal(root);
} } // This code is contributed by Arnab Kundu |
# Python implementation to construct a BST # from its level order traversal import math
# node of a BST class Node:
def __init__(self,data):
self.data = data
self.left = None
self.right = None
# function to get a new node def getNode( data):
# Allocate memory
newNode = Node(data)
# put in the data
newNode.data = data
newNode.left =None
newNode.right = None
return newNode
# function to construct a BST from # its level order traversal def LevelOrder(root , data):
if(root == None):
root = getNode(data)
return root
if(data <= root.data):
root.left = LevelOrder(root.left, data)
else:
root.right = LevelOrder(root.right, data)
return root
def constructBst(arr, n):
if(n == 0):
return None
root = None
for i in range(0, n):
root = LevelOrder(root , arr[i])
return root
# function to print the inorder traversal def inorderTraversal( root):
if (root == None):
return None
inorderTraversal(root.left)
print(root.data,end = " ")
inorderTraversal(root.right)
# Driver program if __name__=='__main__':
arr = [7, 4, 12, 3, 6, 8, 1, 5, 10]
n = len(arr)
root = constructBst(arr, n)
print("Inorder Traversal: ", end = "")
root = inorderTraversal(root)
# This code is contributed by Srathore |
// C# implementation to construct a BST // from its level order traversal using System;
class GFG
{ // node of a BST public class Node
{ public int data;
public Node left, right;
}; // function to get a new node static Node getNode(int data)
{ // Allocate memory
Node newNode = new Node();
// put in the data
newNode.data = data;
newNode.left = newNode.right = null;
return newNode;
} // function to construct a BST from // its level order traversal static Node LevelOrder(Node root,
int data)
{ if(root == null)
{
root = getNode(data);
return root;
}
if(data <= root.data)
root.left = LevelOrder(root.left, data);
else
root.right = LevelOrder(root.right, data);
return root;
} static Node constructBst(int []arr, int n)
{ if(n == 0) return null;
Node root = null;
for(int i = 0; i < n; i++)
root = LevelOrder(root, arr[i]);
return root;
} // function to print the inorder traversal static void inorderTraversal(Node root)
{ if (root == null)
return;
inorderTraversal(root.left);
Console.Write( root.data + " ");
inorderTraversal(root.right);
} // Driver code public static void Main(String []args)
{ int []arr = {7, 4, 12, 3,
6, 8, 1, 5, 10};
int n = arr.Length;
Node root = constructBst(arr, n);
Console.Write("Inorder Traversal: ");
inorderTraversal(root);
} } // This code is contributed by Rajput-Ji |
Output:
Inorder Traversal: 1 3 4 5 6 7 8 10 12
Time Complexity : O(n2)
This is because the above program is like we are inserting n nodes in a bst which takes O(n2) time in the worst case.
This article is contributed by Nishant Balayan. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.