Dada una array ordenada. Escriba una función que cree un árbol de búsqueda binaria balanceada utilizando elementos de array.
Ejemplos:
Input: Array {1, 2, 3}
Output: A Balanced BST
2
/ \
1 3
Input: Array {1, 2, 3, 4}
Output: A Balanced BST
3
/ \
2 4
/
1
Algoritmo: en la publicación anterior , discutimos la construcción de BST a partir de una lista enlazada ordenada. Construir a partir de una array ordenada en tiempo O(n) es más simple ya que podemos obtener el elemento central en tiempo O(1). El siguiente es un algoritmo simple donde primero encontramos el Node medio de la lista y lo convertimos en la raíz del árbol que se construirá.
1) Get the Middle of the array and make it root.
2) Recursively do same for left half and right half.
a) Get the middle of left half and make it left child of the root
created in step 1.
b) Get the middle of right half and make it right child of the
root created in step 1.
A continuación se muestra la implementación del algoritmo anterior. El código principal que crea Balanced BST está resaltado.
C++
// C++ program to print BST in given range
#include<bits/stdc++.h>
using namespace std;
/* A Binary Tree node */
class TNode
{
public:
int data;
TNode* left;
TNode* right;
};
TNode* newNode(int data);
/* A function that constructs Balanced
Binary Search Tree from a sorted array */
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 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 left and right
pointers. */
TNode* newNode(int data)
{
TNode* node = new TNode();
node->data = data;
node->left = NULL;
node->right = NULL;
return node;
}
/* A utility function to print
preorder traversal of BST */
void preOrder(TNode* node)
{
if (node == NULL)
return;
cout << node->data << " ";
preOrder(node->left);
preOrder(node->right);
}
// Driver Code
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
int n = sizeof(arr) / sizeof(arr[0]);
/* Convert List to BST */
TNode *root = sortedArrayToBST(arr, 0, n-1);
cout << "PreOrder Traversal of constructed BST \n";
preOrder(root);
return 0;
}
// This code is contributed by rathbhupendra
C
#include<stdio.h>
#include<stdlib.h>
/* A Binary Tree node */
struct TNode
{
int data;
struct TNode* left;
struct TNode* right;
};
struct TNode* newNode(int data);
/* A function that constructs Balanced Binary Search 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 left and 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;
}
/* A utility function to print preorder traversal of BST */
void preOrder(struct TNode* node)
{
if (node == NULL)
return;
printf("%d ", node->data);
preOrder(node->left);
preOrder(node->right);
}
/* Driver program to test above functions */
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
int n = sizeof(arr)/sizeof(arr[0]);
/* Convert List to BST */
struct TNode *root = sortedArrayToBST(arr, 0, n-1);
printf("n PreOrder Traversal of constructed BST ");
preOrder(root);
return 0;
}
Java
// Java program to print BST in given range
// A binary tree node
class Node {
int data;
Node left, right;
Node(int d) {
data = d;
left = right = null;
}
}
class BinaryTree {
static Node root;
/* A function that constructs Balanced Binary Search Tree
from a sorted array */
Node 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;
Node node = new Node(arr[mid]);
/* Recursively construct the left subtree and make it
left child of root */
node.left = sortedArrayToBST(arr, start, mid - 1);
/* Recursively construct the right subtree and make it
right child of root */
node.right = sortedArrayToBST(arr, mid + 1, end);
return node;
}
/* A utility function to print preorder traversal of BST */
void preOrder(Node node) {
if (node == null) {
return;
}
System.out.print(node.data + " ");
preOrder(node.left);
preOrder(node.right);
}
public static void main(String[] args) {
BinaryTree tree = new BinaryTree();
int arr[] = new int[]{1, 2, 3, 4, 5, 6, 7};
int n = arr.length;
root = tree.sortedArrayToBST(arr, 0, n - 1);
System.out.println("Preorder traversal of constructed BST");
tree.preOrder(root);
}
}
// This code has been contributed by Mayank Jaiswal
Python
# Python code to convert a sorted array # to a balanced Binary Search Tree # binary tree node class Node: def __init__(self, d): self.data = d self.left = None self.right = None # function to convert sorted array to a # balanced BST # input : sorted array of integers # output: root node of balanced BST def sortedArrayToBST(arr): if not arr: return None # find middle index mid = (len(arr)) // 2 # make the middle element the root root = Node(arr[mid]) # left subtree of root has all # values <arr[mid] root.left = sortedArrayToBST(arr[:mid]) # right subtree of root has all # values >arr[mid] root.right = sortedArrayToBST(arr[mid+1:]) return root # A utility function to print the preorder # traversal of the BST def preOrder(node): if not node: return print node.data, preOrder(node.left) preOrder(node.right) # driver program to test above function """ Constructed balanced BST is 4 / \ 2 6 / \ / \ 1 3 5 7 """ arr = [1, 2, 3, 4, 5, 6, 7] root = sortedArrayToBST(arr) print "PreOrder Traversal of constructed BST ", preOrder(root) # This code is contributed by Ishita Tripathi
C#
using System;
// C# program to print BST in given range
// A binary tree node
public class Node
{
public int data;
public Node left, right;
public Node(int d)
{
data = d;
left = right = null;
}
}
public class BinaryTree
{
public static Node root;
/* A function that constructs Balanced Binary Search Tree
from a sorted array */
public virtual Node 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;
Node node = new Node(arr[mid]);
/* Recursively construct the left subtree and make it
left child of root */
node.left = sortedArrayToBST(arr, start, mid - 1);
/* Recursively construct the right subtree and make it
right child of root */
node.right = sortedArrayToBST(arr, mid + 1, end);
return node;
}
/* A utility function to print preorder traversal of BST */
public virtual void preOrder(Node node)
{
if (node == null)
{
return;
}
Console.Write(node.data + " ");
preOrder(node.left);
preOrder(node.right);
}
public static void Main(string[] args)
{
BinaryTree tree = new BinaryTree();
int[] arr = new int[]{1, 2, 3, 4, 5, 6, 7};
int n = arr.Length;
root = tree.sortedArrayToBST(arr, 0, n - 1);
Console.WriteLine("Preorder traversal of constructed BST");
tree.preOrder(root);
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// JavaScript program to print BST in given range
// A binary tree node
class Node
{
constructor(d)
{
this.data = d;
this.left = null;
this.right = null;
}
}
var root = null;
/* A function that constructs Balanced Binary Search Tree
from a sorted array */
function sortedArrayToBST(arr, start, end)
{
/* Base Case */
if (start > end)
{
return null;
}
/* Get the middle element and make it root */
var mid = parseInt((start + end) / 2);
var node = new Node(arr[mid]);
/* Recursively construct the left subtree and make it
left child of root */
node.left = sortedArrayToBST(arr, start, mid - 1);
/* Recursively construct the right subtree and make it
right child of root */
node.right = sortedArrayToBST(arr, mid + 1, end);
return node;
}
/* A utility function to print preorder traversal of BST */
function preOrder(node)
{
if (node == null)
{
return;
}
document.write(node.data + " ");
preOrder(node.left);
preOrder(node.right);
}
var arr = [1, 2, 3, 4, 5, 6, 7];
var n = arr.length;
root = sortedArrayToBST(arr, 0, n - 1);
document.write("Preorder traversal of constructed BST<br>");
preOrder(root);
</script>
Producción:
Preorder traversal of constructed BST 4 2 1 3 6 5 7
Tree representation of above output:
4
2 6
1 3 5 7
Complejidad de tiempo: O(n)
La siguiente es la relación de recurrencia para sortedArrayToBST().
T(n) = 2T(n/2) + C
T(n) --> Time taken for an array of size n
C --> Constant (Finding middle of array and linking root to left
and right subtrees take constant time)
La recurrencia anterior se puede resolver usando el Teorema del Maestro , ya que cae en el caso 1.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA