Se le proporcionan dos árboles de búsqueda binarios equilibrados, por ejemplo, AVL o Red-Black Tree. Escriba una función que fusione los dos BST balanceados dados en un árbol de búsqueda binario balanceado. Sean m elementos en el primer árbol y n elementos en el otro árbol. Su función de combinación debe tomar el tiempo O (m + n).
En las siguientes soluciones, se supone que los tamaños de los árboles también se dan como entrada. Si no se proporciona el tamaño, entonces podemos obtener el tamaño atravesando el árbol (Ver esto ).
Método 1 (Insertar elementos del primer árbol al segundo):
Tome todos los elementos del primer BST uno por uno e insértelos en el segundo BST. Insertar un elemento en un BST de autoequilibrio lleva tiempo de inicio de sesión (consulte esto ), donde n es el tamaño del BST. Entonces, la complejidad temporal de este método es Log(n) + Log(n+1) … Log(m+n-1). El valor de esta expresión estará entre mLogn y mLog(m+n-1). Como optimización, podemos elegir el árbol más pequeño como primer árbol.
Método 2 (Fusionar recorridos en orden):
- Haga un recorrido en orden del primer árbol y almacene el recorrido en una array temporal arr1[]. Este paso lleva O(m) tiempo.
- Haga un recorrido en orden del segundo árbol y almacene el recorrido en otra array temporal arr2[]. Este paso toma tiempo O(n).
- Las arrays creadas en los pasos 1 y 2 son arrays ordenadas. Combine las dos arrays ordenadas en una array de tamaño m + n. Este paso lleva un tiempo O(m+n).
- Construya un árbol balanceado a partir de la array fusionada utilizando la técnica discutida en esta publicación. Este paso lleva un tiempo O(m+n).
La complejidad temporal de este método es O(m+n), que es mejor que el método 1. Este método requiere un tiempo O(m+n) incluso si los BST de entrada no están equilibrados.
A continuación se muestra la implementación de este método.
C++
// C++ program to Merge Two Balanced Binary Search Trees
#include<bits/stdc++.h>
using namespace std;
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class node
{
public:
int data;
node* left;
node* right;
};
// A utility function to merge two sorted arrays into one
int *merge(int arr1[], int arr2[], int m, int n);
// A helper function that stores inorder
// traversal of a tree in inorder array
void storeInorder(node* node, int inorder[],
int *index_ptr);
/* A function that constructs Balanced
Binary Search Tree from a sorted array
See https://www.geeksforgeeks.org/sorted-array-to-balanced-bst/ */
node* sortedArrayToBST(int arr[], int start, int end);
/* This function merges two balanced
BSTs with roots as root1 and root2.
m and n are the sizes of the trees respectively */
node* mergeTrees(node *root1, node *root2, int m, int n)
{
// Store inorder traversal of
// first tree in an array arr1[]
int *arr1 = new int[m];
int i = 0;
storeInorder(root1, arr1, &i);
// Store inorder traversal of second
// tree in another array arr2[]
int *arr2 = new int[n];
int j = 0;
storeInorder(root2, arr2, &j);
// Merge the two sorted array into one
int *mergedArr = merge(arr1, arr2, m, n);
// Construct a tree from the merged
// array and return root of the tree
return sortedArrayToBST (mergedArr, 0, m + n - 1);
}
/* Helper function that allocates
a new node with the given data and
NULL left and right pointers. */
node* newNode(int data)
{
node* Node = new node();
Node->data = data;
Node->left = NULL;
Node->right = NULL;
return(Node);
}
// A utility function to print inorder
// traversal of a given binary tree
void printInorder(node* node)
{
if (node == NULL)
return;
/* first recur on left child */
printInorder(node->left);
cout << node->data << " ";
/* now recur on right child */
printInorder(node->right);
}
// A utility function to merge
// two sorted arrays into one
int *merge(int arr1[], int arr2[], int m, int n)
{
// mergedArr[] is going to contain result
int *mergedArr = new int[m + n];
int i = 0, j = 0, k = 0;
// Traverse through both arrays
while (i < m && j < n)
{
// Pick the smaller element and put it in mergedArr
if (arr1[i] < arr2[j])
{
mergedArr[k] = arr1[i];
i++;
}
else
{
mergedArr[k] = arr2[j];
j++;
}
k++;
}
// If there are more elements in first array
while (i < m)
{
mergedArr[k] = arr1[i];
i++; k++;
}
// If there are more elements in second array
while (j < n)
{
mergedArr[k] = arr2[j];
j++; k++;
}
return mergedArr;
}
// A helper function that stores inorder
// traversal of a tree rooted with node
void storeInorder(node* node, int inorder[], int *index_ptr)
{
if (node == NULL)
return;
/* first recur on left child */
storeInorder(node->left, inorder, index_ptr);
inorder[*index_ptr] = node->data;
(*index_ptr)++; // increase index for next entry
/* now recur on right child */
storeInorder(node->right, inorder, index_ptr);
}
/* A function that constructs Balanced
// Binary Search Tree from a sorted array
See https://www.geeksforgeeks.org/sorted-array-to-balanced-bst/ */
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 *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;
}
/* Driver code*/
int main()
{
/* Create following tree as first balanced BST
100
/ \
50 300
/ \
20 70
*/
node *root1 = newNode(100);
root1->left = newNode(50);
root1->right = newNode(300);
root1->left->left = newNode(20);
root1->left->right = newNode(70);
/* Create following tree as second balanced BST
80
/ \
40 120
*/
node *root2 = newNode(80);
root2->left = newNode(40);
root2->right = newNode(120);
node *mergedTree = mergeTrees(root1, root2, 5, 3);
cout << "Following is Inorder traversal of the merged tree \n";
printInorder(mergedTree);
return 0;
}
// This code is contributed by rathbhupendra
C
// C program to Merge Two Balanced Binary Search Trees
#include <stdio.h>
#include <stdlib.h>
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
};
// A utility function to merge two sorted arrays into one
int *merge(int arr1[], int arr2[], int m, int n);
// A helper function that stores inorder traversal of a tree in inorder array
void storeInorder(struct node* node, int inorder[], int *index_ptr);
/* A function that constructs Balanced Binary Search Tree from a sorted array
See https://www.geeksforgeeks.org/sorted-array-to-balanced-bst/ */
struct node* sortedArrayToBST(int arr[], int start, int end);
/* This function merges two balanced BSTs with roots as root1 and root2.
m and n are the sizes of the trees respectively */
struct node* mergeTrees(struct node *root1, struct node *root2, int m, int n)
{
// Store inorder traversal of first tree in an array arr1[]
int *arr1 = new int[m];
int i = 0;
storeInorder(root1, arr1, &i);
// Store inorder traversal of second tree in another array arr2[]
int *arr2 = new int[n];
int j = 0;
storeInorder(root2, arr2, &j);
// Merge the two sorted array into one
int *mergedArr = merge(arr1, arr2, m, n);
// Construct a tree from the merged array and return root of the tree
return sortedArrayToBST (mergedArr, 0, m+n-1);
}
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct node* newNode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
// A utility function to print inorder traversal of a given binary tree
void printInorder(struct node* node)
{
if (node == NULL)
return;
/* first recur on left child */
printInorder(node->left);
printf("%d ", node->data);
/* now recur on right child */
printInorder(node->right);
}
// A utility function to merge two sorted arrays into one
int *merge(int arr1[], int arr2[], int m, int n)
{
// mergedArr[] is going to contain result
int *mergedArr = new int[m + n];
int i = 0, j = 0, k = 0;
// Traverse through both arrays
while (i < m && j < n)
{
// Pick the smaller element and put it in mergedArr
if (arr1[i] < arr2[j])
{
mergedArr[k] = arr1[i];
i++;
}
else
{
mergedArr[k] = arr2[j];
j++;
}
k++;
}
// If there are more elements in first array
while (i < m)
{
mergedArr[k] = arr1[i];
i++; k++;
}
// If there are more elements in second array
while (j < n)
{
mergedArr[k] = arr2[j];
j++; k++;
}
return mergedArr;
}
// A helper function that stores inorder traversal of a tree rooted with node
void storeInorder(struct node* node, int inorder[], int *index_ptr)
{
if (node == NULL)
return;
/* first recur on left child */
storeInorder(node->left, inorder, index_ptr);
inorder[*index_ptr] = node->data;
(*index_ptr)++; // increase index for next entry
/* now recur on right child */
storeInorder(node->right, inorder, index_ptr);
}
/* A function that constructs Balanced Binary Search Tree from a sorted array
See https://www.geeksforgeeks.org/sorted-array-to-balanced-bst/ */
struct 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;
struct node *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;
}
/* Driver program to test above functions*/
int main()
{
/* Create following tree as first balanced BST
100
/ \
50 300
/ \
20 70
*/
struct node *root1 = newNode(100);
root1->left = newNode(50);
root1->right = newNode(300);
root1->left->left = newNode(20);
root1->left->right = newNode(70);
/* Create following tree as second balanced BST
80
/ \
40 120
*/
struct node *root2 = newNode(80);
root2->left = newNode(40);
root2->right = newNode(120);
struct node *mergedTree = mergeTrees(root1, root2, 5, 3);
printf ("Following is Inorder traversal of the merged tree \n");
printInorder(mergedTree);
getchar();
return 0;
}
Java
// Java program to Merge Two Balanced Binary Search Trees
import java.io.*;
import java.util.ArrayList;
// A binary tree node
class Node {
int data;
Node left, right;
Node(int d) {
data = d;
left = right = null;
}
}
class BinarySearchTree
{
// Root of BST
Node root;
// Constructor
BinarySearchTree() {
root = null;
}
// Inorder traversal of the tree
void inorder()
{
inorderUtil(this.root);
}
// Utility function for inorder traversal of the tree
void inorderUtil(Node node)
{
if(node==null)
return;
inorderUtil(node.left);
System.out.print(node.data + " ");
inorderUtil(node.right);
}
// A Utility Method that stores inorder traversal of a tree
public ArrayList<Integer> storeInorderUtil(Node node, ArrayList<Integer> list)
{
if(node == null)
return list;
//recur on the left child
storeInorderUtil(node.left, list);
// Adds data to the list
list.add(node.data);
//recur on the right child
storeInorderUtil(node.right, list);
return list;
}
// Method that stores inorder traversal of a tree
ArrayList<Integer> storeInorder(Node node)
{
ArrayList<Integer> list1 = new ArrayList<>();
ArrayList<Integer> list2 = storeInorderUtil(node,list1);
return list2;
}
// Method that merges two ArrayLists into one.
ArrayList<Integer> merge(ArrayList<Integer>list1, ArrayList<Integer>list2, int m, int n)
{
// list3 will contain the merge of list1 and list2
ArrayList<Integer> list3 = new ArrayList<>();
int i=0;
int j=0;
//Traversing through both ArrayLists
while( i<m && j<n)
{
// Smaller one goes into list3
if(list1.get(i)<list2.get(j))
{
list3.add(list1.get(i));
i++;
}
else
{
list3.add(list2.get(j));
j++;
}
}
// Adds the remaining elements of list1 into list3
while(i<m)
{
list3.add(list1.get(i));
i++;
}
// Adds the remaining elements of list2 into list3
while(j<n)
{
list3.add(list2.get(j));
j++;
}
return list3;
}
// Method that converts an ArrayList to a BST
Node ALtoBST(ArrayList<Integer>list, 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(list.get(mid));
/* Recursively construct the left subtree and make it
left child of root */
node.left = ALtoBST(list, start, mid-1);
/* Recursively construct the right subtree and make it
right child of root */
node.right = ALtoBST(list, mid+1, end);
return node;
}
// Method that merges two trees into a single one.
Node mergeTrees(Node node1, Node node2)
{
//Stores Inorder of tree1 to list1
ArrayList<Integer>list1 = storeInorder(node1);
//Stores Inorder of tree2 to list2
ArrayList<Integer>list2 = storeInorder(node2);
// Merges both list1 and list2 into list3
ArrayList<Integer>list3 = merge(list1, list2, list1.size(), list2.size());
//Eventually converts the merged list into resultant BST
Node node = ALtoBST(list3, 0, list3.size()-1);
return node;
}
// Driver function
public static void main (String[] args)
{
/* Creating following tree as First balanced BST
100
/ \
50 300
/ \
20 70
*/
BinarySearchTree tree1 = new BinarySearchTree();
tree1.root = new Node(100);
tree1.root.left = new Node(50);
tree1.root.right = new Node(300);
tree1.root.left.left = new Node(20);
tree1.root.left.right = new Node(70);
/* Creating following tree as second balanced BST
80
/ \
40 120
*/
BinarySearchTree tree2 = new BinarySearchTree();
tree2.root = new Node(80);
tree2.root.left = new Node(40);
tree2.root.right = new Node(120);
BinarySearchTree tree = new BinarySearchTree();
tree.root = tree.mergeTrees(tree1.root, tree2.root);
System.out.println("The Inorder traversal of the merged BST is: ");
tree.inorder();
}
}
// This code has been contributed by Kamal Rawal
Python3
# A binary tree node has data, pointer to left child
# and a pointer to right child
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
# A utility function to merge two sorted arrays into one
# Time Complexity of below function: O(m + n)
# Space Complexity of below function: O(m + n)
def merge_sorted_arr(arr1, arr2):
arr = []
i = j = 0
while i < len(arr1) and j < len(arr2):
if arr1[i] <= arr2[j]:
arr.append(arr1[i])
i += 1
else:
arr.append(arr2[j])
j += 1
while i < len(arr1):
arr.append(arr1[i])
i += 1
while i < len(arr2):
arr.append(arr2[j])
j += 1
return arr
# A helper function that stores inorder
# traversal of a tree in arr
def inorder(root, arr = []):
if root:
inorder(root.left, arr)
arr.append(root.val)
inorder(root.right, arr)
# A utility function to insert the values
# in the individual Tree
def insert(root, val):
if not root:
return Node(val)
if root.val == val:
return root
elif root.val > val:
root.left = insert(root.left, val)
else:
root.right = insert(root.right, val)
return root
# Converts the merged array to a balanced BST
# Explanation of the below code:
# https://www.geeksforgeeks.org/sorted-array-to-balanced-bst/
def arr_to_bst(arr):
if not arr:
return None
mid = len(arr) // 2
root = Node(arr[mid])
root.left = arr_to_bst(arr[:mid])
root.right = arr_to_bst(arr[mid + 1:])
return root
if __name__=='__main__':
root1 = root2 = None
# Inserting values in first tree
root1 = insert(root1, 100)
root1 = insert(root1, 50)
root1 = insert(root1, 300)
root1 = insert(root1, 20)
root1 = insert(root1, 70)
# Inserting values in second tree
root2 = insert(root2, 80)
root2 = insert(root2, 40)
root2 = insert(root2, 120)
arr1 = []
inorder(root1, arr1)
arr2 = []
inorder(root2, arr2)
arr = merge_sorted_arr(arr1, arr2)
root = arr_to_bst(arr)
res = []
inorder(root, res)
print('Following is Inorder traversal of the merged tree')
for i in res:
print(i, end = ' ')
# This code is contributed by Flarow4
C#
// C# program to Merge Two Balanced Binary Search Trees
using System;
using System.Collections.Generic;
// 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 BinarySearchTree
{
// Root of BST
public Node root;
// Constructor
public BinarySearchTree()
{
root = null;
}
// Inorder traversal of the tree
public virtual void inorder()
{
inorderUtil(this.root);
}
// Utility function for inorder traversal of the tree
public virtual void inorderUtil(Node node)
{
if (node == null)
{
return;
}
inorderUtil(node.left);
Console.Write(node.data + " ");
inorderUtil(node.right);
}
// A Utility Method that stores inorder traversal of a tree
public virtual List<int> storeInorderUtil(Node node, List<int> list)
{
if (node == null)
{
return list;
}
//recur on the left child
storeInorderUtil(node.left, list);
// Adds data to the list
list.Add(node.data);
//recur on the right child
storeInorderUtil(node.right, list);
return list;
}
// Method that stores inorder traversal of a tree
public virtual List<int> storeInorder(Node node)
{
List<int> list1 = new List<int>();
List<int> list2 = storeInorderUtil(node,list1);
return list2;
}
// Method that merges two ArrayLists into one.
public virtual List<int> merge(List<int> list1, List<int> list2, int m, int n)
{
// list3 will contain the merge of list1 and list2
List<int> list3 = new List<int>();
int i = 0;
int j = 0;
//Traversing through both ArrayLists
while (i < m && j < n)
{
// Smaller one goes into list3
if (list1[i] < list2[j])
{
list3.Add(list1[i]);
i++;
}
else
{
list3.Add(list2[j]);
j++;
}
}
// Adds the remaining elements of list1 into list3
while (i < m)
{
list3.Add(list1[i]);
i++;
}
// Adds the remaining elements of list2 into list3
while (j < n)
{
list3.Add(list2[j]);
j++;
}
return list3;
}
// Method that converts an ArrayList to a BST
public virtual Node ALtoBST(List<int> list, 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(list[mid]);
/* Recursively construct the left subtree and make it
left child of root */
node.left = ALtoBST(list, start, mid - 1);
/* Recursively construct the right subtree and make it
right child of root */
node.right = ALtoBST(list, mid + 1, end);
return node;
}
// Method that merges two trees into a single one.
public virtual Node mergeTrees(Node node1, Node node2)
{
//Stores Inorder of tree1 to list1
List<int> list1 = storeInorder(node1);
//Stores Inorder of tree2 to list2
List<int> list2 = storeInorder(node2);
// Merges both list1 and list2 into list3
List<int> list3 = merge(list1, list2, list1.Count, list2.Count);
//Eventually converts the merged list into resultant BST
Node node = ALtoBST(list3, 0, list3.Count - 1);
return node;
}
// Driver function
public static void Main(string[] args)
{
/* Creating following tree as First balanced BST
100
/ \
50 300
/ \
20 70
*/
BinarySearchTree tree1 = new BinarySearchTree();
tree1.root = new Node(100);
tree1.root.left = new Node(50);
tree1.root.right = new Node(300);
tree1.root.left.left = new Node(20);
tree1.root.left.right = new Node(70);
/* Creating following tree as second balanced BST
80
/ \
40 120
*/
BinarySearchTree tree2 = new BinarySearchTree();
tree2.root = new Node(80);
tree2.root.left = new Node(40);
tree2.root.right = new Node(120);
BinarySearchTree tree = new BinarySearchTree();
tree.root = tree.mergeTrees(tree1.root, tree2.root);
Console.WriteLine("The Inorder traversal of the merged BST is: ");
tree.inorder();
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// JavaScript program to Merge Two
// Balanced Binary Search Trees
// A binary tree node
class Node {
constructor(d) {
this.data = d;
this.left = null;
this.right = null;
}
}
class BinarySearchTree {
// Constructor
constructor() {
this.root = null;
}
// Inorder traversal of the tree
inorder() {
this.inorderUtil(this.root);
}
// Utility function for inorder traversal of the tree
inorderUtil(node) {
if (node == null) {
return;
}
this.inorderUtil(node.left);
document.write(node.data + " ");
this.inorderUtil(node.right);
}
// A Utility Method that stores
// inorder traversal of a tree
storeInorderUtil(node, list) {
if (node == null) {
return list;
}
//recur on the left child
this.storeInorderUtil(node.left, list);
// Adds data to the list
list.push(node.data);
//recur on the right child
this.storeInorderUtil(node.right, list);
return list;
}
// Method that stores inorder traversal of a tree
storeInorder(node) {
var list1 = [];
var list2 = this.storeInorderUtil(node, list1);
return list2;
}
// Method that merges two ArrayLists into one.
merge(list1, list2, m, n) {
// list3 will contain the merge of list1 and list2
var list3 = [];
var i = 0;
var j = 0;
//Traversing through both ArrayLists
while (i < m && j < n) {
// Smaller one goes into list3
if (list1[i] < list2[j]) {
list3.push(list1[i]);
i++;
} else {
list3.push(list2[j]);
j++;
}
}
// Adds the remaining elements of list1 into list3
while (i < m) {
list3.push(list1[i]);
i++;
}
// Adds the remaining elements of list2 into list3
while (j < n) {
list3.push(list2[j]);
j++;
}
return list3;
}
// Method that converts an ArrayList to a BST
ALtoBST(list, 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(list[mid]);
/* Recursively construct the left subtree and make it
left child of root */
node.left = this.ALtoBST(list, start, mid - 1);
/* Recursively construct the right subtree and make it
right child of root */
node.right = this.ALtoBST(list, mid + 1, end);
return node;
}
// Method that merges two trees into a single one.
mergeTrees(node1, node2) {
//Stores Inorder of tree1 to list1
var list1 = this.storeInorder(node1);
//Stores Inorder of tree2 to list2
var list2 = this.storeInorder(node2);
// Merges both list1 and list2 into list3
var list3 =
this.merge(list1, list2, list1.length, list2.length);
//Eventually converts the merged list into resultant BST
var node = this.ALtoBST(list3, 0, list3.length - 1);
return node;
}
}
// Driver function
/* Creating following tree as First balanced BST
100
/ \
50 300
/ \
20 70
*/
var tree1 = new BinarySearchTree();
tree1.root = new Node(100);
tree1.root.left = new Node(50);
tree1.root.right = new Node(300);
tree1.root.left.left = new Node(20);
tree1.root.left.right = new Node(70);
/* Creating following tree as second balanced BST
80
/ \
40 120
*/
var tree2 = new BinarySearchTree();
tree2.root = new Node(80);
tree2.root.left = new Node(40);
tree2.root.right = new Node(120);
var tree = new BinarySearchTree();
tree.root = tree.mergeTrees(tree1.root, tree2.root);
document.write(
"Following is Inorder traversal of the merged tree <br>"
);
tree.inorder();
</script>
Producción:
Following is Inorder traversal of the merged tree 20 40 50 70 80 100 120 300
Método 3 (Fusión local mediante DLL):
Podemos usar una lista doblemente enlazada para fusionar árboles en su lugar. Los siguientes son los pasos.
- Convierta los dos árboles de búsqueda binarios dados en una lista doblemente vinculada en su lugar (consulte esta publicación para este paso).
- Combine las dos listas vinculadas ordenadas (consulte esta publicación para este paso).
- Cree un árbol de búsqueda binario equilibrado a partir de la lista fusionada creada en el paso 2. (Consulte esta publicación para este paso)
La complejidad temporal de este método también es O(m+n) y este método realiza la conversión en el lugar.
Gracias a Dheeraj y Ronzii por sugerir este método.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA