Dados dos árboles binarios. Necesitamos fusionarlos en un nuevo árbol binario. La regla de fusión es que si dos Nodes se superponen, entonces suma los valores de los Nodes como el nuevo valor del Node fusionado. De lo contrario, el Node no nulo se utilizará como Node del nuevo árbol.
Ejemplo:
Input:
Tree 1 Tree 2
2 3
/ \ / \
1 4 6 1
/ \ \
5 2 7
Output: Merged tree:
5
/ \
7 5
/ \ \
5 2 7
Nota: El proceso de fusión debe comenzar desde los Nodes raíz de ambos árboles.
Algoritmo recursivo:
- Atraviesa el árbol en modo de pedido anticipado
- Compruebe si ambos Nodes del árbol son NULL
- Si no, entonces actualice el valor
- Repetir para subárboles izquierdos
- Repetir para subárboles derechos
- Devolver la raíz del árbol actualizado
C++
// C++ program to Merge Two Binary Trees
#include <bits/stdc++.h>
using namespace std;
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node
{
int data;
struct Node *left, *right;
};
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
Node *newNode(int data)
{
Node *new_node = new Node;
new_node->data = data;
new_node->left = new_node->right = NULL;
return new_node;
}
/* Given a binary tree, print its nodes in inorder*/
void inorder(Node * node)
{
if (!node)
return;
/* first recur on left child */
inorder(node->left);
/* then print the data of node */
cout<<node->data<<" ";
/* now recur on right child */
inorder(node->right);
}
/* Function to merge given two binary trees*/
Node *MergeTrees(Node * t1, Node * t2)
{
if (!t1)
return t2;
if (!t2)
return t1;
t1->data += t2->data;
t1->left = MergeTrees(t1->left, t2->left);
t1->right = MergeTrees(t1->right, t2->right);
return t1;
}
// Driver code
int main()
{
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
Node *root1 = newNode(1);
root1->left = newNode(2);
root1->right = newNode(3);
root1->left->left = newNode(4);
root1->left->right = newNode(5);
root1->right->right = newNode(6);
/* Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
Node *root2 = newNode(4);
root2->left = newNode(1);
root2->right = newNode(7);
root2->left->left = newNode(3);
root2->right->left = newNode(2);
root2->right->right = newNode(6);
Node *root3 = MergeTrees(root1, root2);
printf("The Merged Binary Tree is:\n");
inorder(root3);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
C
// C program to Merge Two Binary Trees
#include<stdio.h>
#include<stdlib.h>
/* A binary tree node has data, pointer to left child
and a pointer to right child */
typedef struct Node
{
int data;
struct Node *left, *right;
}Node;
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
Node *newNode(int data)
{
Node *new_node = (Node *)malloc(sizeof(Node));
new_node->data = data;
new_node->left = new_node->right = NULL;
return new_node;
}
/* Given a binary tree, print its nodes in inorder*/
void inorder(Node * node)
{
if (!node)
return;
/* first recur on left child */
inorder(node->left);
/* then print the data of node */
printf("%d ", node->data);
/* now recur on right child */
inorder(node->right);
}
/* Function to merge given two binary trees*/
Node *MergeTrees(Node * t1, Node * t2)
{
if (!t1)
return t2;
if (!t2)
return t1;
t1->data += t2->data;
t1->left = MergeTrees(t1->left, t2->left);
t1->right = MergeTrees(t1->right, t2->right);
return t1;
}
// Driver code
int main()
{
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
Node *root1 = newNode(1);
root1->left = newNode(2);
root1->right = newNode(3);
root1->left->left = newNode(4);
root1->left->right = newNode(5);
root1->right->right = newNode(6);
/* Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
Node *root2 = newNode(4);
root2->left = newNode(1);
root2->right = newNode(7);
root2->left->left = newNode(3);
root2->right->left = newNode(2);
root2->right->right = newNode(6);
Node *root3 = MergeTrees(root1, root2);
printf("The Merged Binary Tree is:\n");
inorder(root3);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
// Java program to Merge Two Binary Trees
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class Node
{
int data;
Node left, right;
public Node(int data, Node left, Node right) {
this.data = data;
this.left = left;
this.right = right;
}
/* Helper method that allocates a new node with the
given data and NULL left and right pointers. */
static Node newNode(int data)
{
return new Node(data, null, null);
}
/* Given a binary tree, print its nodes in inorder*/
static void inorder(Node node)
{
if (node == null)
return;
/* first recur on left child */
inorder(node.left);
/* then print the data of node */
System.out.printf("%d ", node.data);
/* now recur on right child */
inorder(node.right);
}
/* Method to merge given two binary trees*/
static Node MergeTrees(Node t1, Node t2)
{
if (t1 == null)
return t2;
if (t2 == null)
return t1;
t1.data += t2.data;
t1.left = MergeTrees(t1.left, t2.left);
t1.right = MergeTrees(t1.right, t2.right);
return t1;
}
// Driver method
public static void main(String[] args)
{
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
Node root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
root1.right.right = newNode(6);
/* Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
Node root2 = newNode(4);
root2.left = newNode(1);
root2.right = newNode(7);
root2.left.left = newNode(3);
root2.right.left = newNode(2);
root2.right.right = newNode(6);
Node root3 = MergeTrees(root1, root2);
System.out.printf("The Merged Binary Tree is:\n");
inorder(root3);
}
}
// This code is contributed by Gaurav Miglani
Python3
# Python3 program to Merge Two Binary Trees
# Helper class that allocates a new node
# with the given data and None left and
# right pointers.
class newNode:
def __init__(self, data):
self.data = data
self.left = self.right = None
# Given a binary tree, prints nodes
# in inorder
def inorder(node):
if (not node):
return
# first recur on left child
inorder(node.left)
# then print the data of node
print(node.data, end = " ")
# now recur on right child
inorder(node.right)
# Function to merge given two
# binary trees
def MergeTrees(t1, t2):
if (not t1):
return t2
if (not t2):
return t1
t1.data += t2.data
t1.left = MergeTrees(t1.left, t2.left)
t1.right = MergeTrees(t1.right, t2.right)
return t1
# Driver code
if __name__ == '__main__':
# Let us construct the first Binary Tree
# 1
# / \
# 2 3
# / \ \
# 4 5 6
root1 = newNode(1)
root1.left = newNode(2)
root1.right = newNode(3)
root1.left.left = newNode(4)
root1.left.right = newNode(5)
root1.right.right = newNode(6)
# Let us construct the second Binary Tree
# 4
# / \
# 1 7
# / / \
# 3 2 6
root2 = newNode(4)
root2.left = newNode(1)
root2.right = newNode(7)
root2.left.left = newNode(3)
root2.right.left = newNode(2)
root2.right.right = newNode(6)
root3 = MergeTrees(root1, root2)
print("The Merged Binary Tree is:")
inorder(root3)
# This code is contributed by PranchalK
C#
// C# program to Merge Two Binary Trees
using System;
/* A binary tree node has data, pointer
to left child and a pointer to right child */
public class Node
{
public int data;
public Node left, right;
public Node(int data, Node left,
Node right)
{
this.data = data;
this.left = left;
this.right = right;
}
/* Helper method that allocates a new
node with the given data and NULL left
and right pointers. */
public static Node newNode(int data)
{
return new Node(data, null, null);
}
/* Given a binary tree, print its
nodes in inorder*/
public static void preorder(Node node)
{
if (node == null)
{
return;
}
/* then print the data of node */
Console.Write("{0:D} ", node.data);
/* first recur on left child */
inorder(node.left);
/* now recur on right child */
inorder(node.right);
}
/* Method to merge given two binary trees*/
public static Node MergeTrees(Node t1, Node t2)
{
if (t1 == null)
{
return t2;
}
if (t2 == null)
{
return t1;
}
t1.data += t2.data;
t1.left = MergeTrees(t1.left, t2.left);
t1.right = MergeTrees(t1.right, t2.right);
return t1;
}
// Driver Code
public static void Main(string[] args)
{
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
Node root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
root1.right.right = newNode(6);
/* Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
Node root2 = newNode(4);
root2.left = newNode(1);
root2.right = newNode(7);
root2.left.left = newNode(3);
root2.right.left = newNode(2);
root2.right.right = newNode(6);
Node root3 = MergeTrees(root1, root2);
Console.Write("The Merged Binary Tree is:\n");
preorder(root3);
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// Javascript program to Merge Two Binary Trees
class Node
{
constructor(data)
{
this.left = null;
this.right = null;
this.data = data;
}
}
// Helper method that allocates a new
// node with the given data and NULL
// left and right pointers.
function newNode(data)
{
return new Node(data);
}
// Given a binary tree, print its
// nodes in inorder
function inorder(node)
{
if (node == null)
return;
// First recur on left child
inorder(node.left);
// Then print the data of node
document.write(node.data + " ");
// Now recur on right child
inorder(node.right);
}
// Method to merge given two binary trees
function MergeTrees(t1, t2)
{
if (t1 == null)
return t2;
if (t2 == null)
return t1;
t1.data += t2.data;
t1.left = MergeTrees(t1.left, t2.left);
t1.right = MergeTrees(t1.right, t2.right);
return t1;
}
// Driver code
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
let root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
root1.right.right = newNode(6);
/* Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
let root2 = newNode(4);
root2.left = newNode(1);
root2.right = newNode(7);
root2.left.left = newNode(3);
root2.right.left = newNode(2);
root2.right.right = newNode(6);
let root3 = MergeTrees(root1, root2);
document.write("The Merged Binary Tree is:" + "</br>");
inorder(root3);
// This code is contributed by divyeshrabadiya07
</script>
Producción
The Merged Binary Tree is: 7 3 5 5 2 10 12
Análisis de Complejidad:
- Complejidad temporal: O(n)
Se necesita atravesar un total de n Nodes. Aquí, n representa el número mínimo de Nodes de los dos árboles dados. - Espacio auxiliar: O(n)
La profundidad del árbol de recurrencia puede llegar hasta n en el caso de un árbol sesgado. En el caso promedio, la profundidad será O (logn).
Algoritmo iterativo:
- crear una pila
- Empuje los Nodes raíz de ambos árboles en la pila.
- Si bien la pila no está vacía, realice los siguientes pasos:
- Extraiga un par de Nodes de la parte superior de la pila
- Por cada par de Nodes eliminado, agregue los valores correspondientes a los dos Nodes y actualice el valor del Node correspondiente en el primer árbol
- Si el hijo izquierdo del primer árbol existe, empuje el hijo izquierdo (par) de ambos árboles a la pila.
- Si el hijo izquierdo del primer árbol no existe, agregue el hijo izquierdo del segundo árbol al Node actual del primer árbol
- Haga lo mismo para el par de niños correctos también.
- Si ambos Nodes actuales son NULOS, continúe extrayendo los siguientes Nodes de la pila.
- Devolver la raíz del árbol actualizado
Implementación:
C++
// C++ program to Merge Two Binary Trees
#include <bits/stdc++.h>
using namespace std;
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node
{
int data;
struct Node *left, *right;
};
// Structure to store node pair onto stack
struct snode
{
Node *l, *r;
};
/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
Node *newNode(int data)
{
Node *new_node = new Node;
new_node->data = data;
new_node->left = new_node->right = NULL;
return new_node;
}
/* Given a binary tree, print its nodes in inorder*/
void inorder(Node * node)
{
if (! node)
return;
/* first recur on left child */
inorder(node->left);
/* then print the data of node */
printf("%d ", node->data);
/* now recur on right child */
inorder(node->right);
}
/* Function to merge given two binary trees*/
Node* MergeTrees(Node* t1, Node* t2)
{
if (! t1)
return t2;
if (! t2)
return t1;
stack<snode> s;
snode temp;
temp.l = t1;
temp.r = t2;
s.push(temp);
snode n;
while (! s.empty())
{
n = s.top();
s.pop();
if (n.l == NULL|| n.r == NULL)
continue;
n.l->data += n.r->data;
if (n.l->left == NULL)
n.l->left = n.r->left;
else
{
snode t;
t.l = n.l->left;
t.r = n.r->left;
s.push(t);
}
if (n.l->right == NULL)
n.l->right = n.r->right;
else
{
snode t;
t.l = n.l->right;
t.r = n.r->right;
s.push(t);
}
}
return t1;
}
// Driver code
int main()
{
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
Node *root1 = newNode(1);
root1->left = newNode(2);
root1->right = newNode(3);
root1->left->left = newNode(4);
root1->left->right = newNode(5);
root1->right->right = newNode(6);
/* Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
Node *root2 = newNode(4);
root2->left = newNode(1);
root2->right = newNode(7);
root2->left->left = newNode(3);
root2->right->left = newNode(2);
root2->right->right = newNode(6);
Node *root3 = MergeTrees(root1, root2);
printf("The Merged Binary Tree is:\n");
inorder(root3);
return 0;
}
Java
// Java program to Merge Two Binary Trees
import java.util.*;
class GFG{
/* A binary tree node has data, pointer to left child
and a pointer to right child */
static class Node
{
int data;
Node left, right;
};
// Structure to store node pair onto stack
static class snode
{
Node l, r;
};
/* Helper function that allocates a new node with the
given data and null left and right pointers. */
static Node newNode(int data)
{
Node new_node = new Node();
new_node.data = data;
new_node.left = new_node.right = null;
return new_node;
}
/* Given a binary tree, print its nodes in inorder*/
static void inorder(Node node)
{
if (node == null)
return;
/* first recur on left child */
inorder(node.left);
/* then print the data of node */
System.out.printf("%d ", node.data);
/* now recur on right child */
inorder(node.right);
}
/* Function to merge given two binary trees*/
static Node MergeTrees(Node t1, Node t2)
{
if ( t1 == null)
return t2;
if ( t2 == null)
return t1;
Stack<snode> s = new Stack<>();
snode temp = new snode();
temp.l = t1;
temp.r = t2;
s.add(temp);
snode n;
while (! s.isEmpty())
{
n = s.peek();
s.pop();
if (n.l == null|| n.r == null)
continue;
n.l.data += n.r.data;
if (n.l.left == null)
n.l.left = n.r.left;
else
{
snode t = new snode();
t.l = n.l.left;
t.r = n.r.left;
s.add(t);
}
if (n.l.right == null)
n.l.right = n.r.right;
else
{
snode t = new snode();
t.l = n.l.right;
t.r = n.r.right;
s.add(t);
}
}
return t1;
}
// Driver code
public static void main(String[] args)
{
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
Node root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
root1.right.right = newNode(6);
/* Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
Node root2 = newNode(4);
root2.left = newNode(1);
root2.right = newNode(7);
root2.left.left = newNode(3);
root2.right.left = newNode(2);
root2.right.right = newNode(6);
Node root3 = MergeTrees(root1, root2);
System.out.printf("The Merged Binary Tree is:\n");
inorder(root3);
}
}
// This code is contributed by gauravrajput1
Python3
# Python3 program to Merge Two Binary Trees
''' A binary tree node has data, pointer to left child
and a pointer to right child '''
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Structure to store node pair onto stack
class snode:
def __init__(self, l, r):
self.l = l
self.r = r
''' Helper function that allocates a new node with the
given data and None left and right pointers. '''
def newNode(data):
new_node = Node(data)
return new_node
''' Given a binary tree, print its nodes in inorder'''
def inorder(node):
if (not node):
return;
''' first recur on left child '''
inorder(node.left);
''' then print the data of node '''
print(node.data, end=' ');
''' now recur on right child '''
inorder(node.right);
''' Function to merge given two binary trees'''
def MergeTrees(t1, t2):
if (not t1):
return t2;
if (not t2):
return t1;
s = []
temp = snode(t1, t2)
s.append(temp);
n = None
while (len(s) != 0):
n = s[-1]
s.pop();
if (n.l == None or n.r == None):
continue;
n.l.data += n.r.data;
if (n.l.left == None):
n.l.left = n.r.left;
else:
t=snode(n.l.left, n.r.left)
s.append(t);
if (n.l.right == None):
n.l.right = n.r.right;
else:
t=snode(n.l.right, n.r.right)
s.append(t);
return t1;
# Driver code
if __name__=='__main__':
''' Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
'''
root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
root1.right.right = newNode(6);
''' Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 '''
root2 = newNode(4);
root2.left = newNode(1);
root2.right = newNode(7);
root2.left.left = newNode(3);
root2.right.left = newNode(2);
root2.right.right = newNode(6);
root3 = MergeTrees(root1, root2);
print("The Merged Binary Tree is:");
inorder(root3);
# This code is contributed by rutvik76
C#
// C# program to Merge Two Binary Trees
using System;
using System.Collections.Generic;
class GFG{
// A binary tree node has data, pointer
// to left child and a pointer to right
// child
public class Node
{
public int data;
public Node left, right;
};
// Structure to store node pair onto stack
public class snode
{
public Node l, r;
};
// Helper function that allocates a new
// node with the given data and null
// left and right pointers.
static Node newNode(int data)
{
Node new_node = new Node();
new_node.data = data;
new_node.left = new_node.right = null;
return new_node;
}
// Given a binary tree, print its
// nodes in inorder
static void inorder(Node node)
{
if (node == null)
return;
// First recur on left child
inorder(node.left);
// Then print the data of node
Console.Write(node.data + " ");
// Now recur on right child
inorder(node.right);
}
// Function to merge given two binary trees
static Node MergeTrees(Node t1, Node t2)
{
if ( t1 == null)
return t2;
if ( t2 == null)
return t1;
Stack<snode> s = new Stack<snode>();
snode temp = new snode();
temp.l = t1;
temp.r = t2;
s.Push(temp);
snode n;
while (s.Count != 0)
{
n = s.Peek();
s.Pop();
if (n.l == null|| n.r == null)
continue;
n.l.data += n.r.data;
if (n.l.left == null)
n.l.left = n.r.left;
else
{
snode t = new snode();
t.l = n.l.left;
t.r = n.r.left;
s.Push(t);
}
if (n.l.right == null)
n.l.right = n.r.right;
else
{
snode t = new snode();
t.l = n.l.right;
t.r = n.r.right;
s.Push(t);
}
}
return t1;
}
// Driver code
public static void Main(String[] args)
{
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
Node root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
root1.right.right = newNode(6);
/* Let us construct the second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
Node root2 = newNode(4);
root2.left = newNode(1);
root2.right = newNode(7);
root2.left.left = newNode(3);
root2.right.left = newNode(2);
root2.right.right = newNode(6);
Node root3 = MergeTrees(root1, root2);
Console.Write("The Merged Binary Tree is:\n");
inorder(root3);
}
}
// This code is contributed by aashish1995
Javascript
<script>
// JavaScript program to Merge Two Binary Trees
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class Node
{
constructor()
{
this.data=0;
this.left=this.right=null;
}
}
// Structure to store node pair onto stack
class snode
{
constructor()
{
this.l=null;
this.r=null;
}
}
/* Helper function that allocates a new node with the
given data and null left and right pointers. */
function newNode(data)
{
let new_node = new Node();
new_node.data = data;
new_node.left = new_node.right = null;
return new_node;
}
/* Given a binary tree, print its nodes in inorder*/
function inorder(node)
{
if (node == null)
return;
/* first recur on left child */
inorder(node.left);
/* then print the data of node */
document.write(node.data+" ");
/* now recur on right child */
inorder(node.right);
}
/* Function to merge given two binary trees*/
function MergeTrees(t1,t2)
{
if ( t1 == null)
return t2;
if ( t2 == null)
return t1;
let s = [];
let temp = new snode();
temp.l = t1;
temp.r = t2;
s.push(temp);
let n;
while ( s.length!=0)
{
n = s.pop();
if (n.l == null|| n.r == null)
continue;
n.l.data += n.r.data;
if (n.l.left == null)
n.l.left = n.r.left;
else
{
let t = new snode();
t.l = n.l.left;
t.r = n.r.left;
s.push(t);
}
if (n.l.right == null)
n.l.right = n.r.right;
else
{
let t = new snode();
t.l = n.l.right;
t.r = n.r.right;
s.push(t);
}
}
return t1;
}
// Driver code
/* Let us construct the first Binary Tree
1
/ \
2 3
/ \ \
4 5 6
*/
let root1 = newNode(1);
root1.left = newNode(2);
root1.right = newNode(3);
root1.left.left = newNode(4);
root1.left.right = newNode(5);
root1.right.right = newNode(6);
/* Let us construct second Binary Tree
4
/ \
1 7
/ / \
3 2 6 */
let root2 = newNode(4);
root2.left = newNode(1);
root2.right = newNode(7);
root2.left.left = newNode(3);
root2.right.left = newNode(2);
root2.right.right = newNode(6);
let root3 = MergeTrees(root1, root2);
document.write("The Merged Binary Tree is:<br>");
inorder(root3);
// This code is contributed by unknown2108
</script>
Producción
The Merged Binary Tree is: 7 3 5 5 2 10 12
Análisis de Complejidad:
- Complejidad temporal: O(n)
Se necesita atravesar un total de n Nodes. Aquí, n representa el número mínimo de Nodes de los dos árboles dados. - Espacio auxiliar: O(n)
La profundidad de la pila puede llegar hasta n en caso de un árbol sesgado.
Este artículo es una contribución de Aakash Pal . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA