Suma de todos los Nodes límite de un árbol binario

Posted on by Rudeus Greyrat

Dado un árbol binario, la tarea es imprimir la suma de todos los Nodes límite del árbol.
 

Ejemplos: 

Input:
               1
             /   \
            2     3
           / \   / \
          4   5 6   7
Output: 28

Input:
                1
              /   \
             2     3
              \    /
               4  5
                  \
                   6
                  / \
                 7   8
Output: 36

Enfoque: Ya hemos discutido el cruce de límites de un árbol binario . Aquí encontraremos la suma de los Nodes límite del árbol binario dado en cuatro pasos: 

  • Sumar todos los Nodes del límite izquierdo,
  • Sume todos los Nodes hoja del subárbol izquierdo,
  • Sume todos los Nodes hoja del subárbol derecho y
  • Sume todos los Nodes del límite derecho.

Tendremos que cuidar una cosa para que los Nodes no se vuelvan a sumar, es decir, el Node más a la izquierda es también el Node hoja del árbol.

A continuación se muestra la implementación del enfoque anterior: 

C++

// C++ implementation of the approach
#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;
    struct Node* right;
};
 
// Utility function to create a node
Node* newNode(int data)
{
    Node* temp = new Node;
 
    temp->left = NULL;
    temp->right = NULL;
    temp->data = data;
 
    return temp;
}
 
// Function to sum up all the left boundary nodes
// except the leaf nodes
void LeftBoundary(Node* root, int& sum_of_boundary_nodes)
{
    if (root) {
        if (root->left) {
            sum_of_boundary_nodes += root->data;
            LeftBoundary(root->left, sum_of_boundary_nodes);
        }
        else if (root->right) {
            sum_of_boundary_nodes += root->data;
            LeftBoundary(root->right, sum_of_boundary_nodes);
        }
    }
}
 
// Function to sum up all the right boundary nodes
// except the leaf nodes
void RightBoundary(Node* root, int& sum_of_boundary_nodes)
{
    if (root) {
        if (root->right) {
            RightBoundary(root->right, sum_of_boundary_nodes);
            sum_of_boundary_nodes += root->data;
        }
        else if (root->left) {
            RightBoundary(root->left, sum_of_boundary_nodes);
            sum_of_boundary_nodes += root->data;
        }
    }
}
 
// Function to sum up all the leaf nodes
// of a binary tree
void Leaves(Node* root, int& sum_of_boundary_nodes)
{
    if (root) {
        Leaves(root->left, sum_of_boundary_nodes);
 
        // Sum it up if it is a leaf node
        if (!(root->left) && !(root->right))
            sum_of_boundary_nodes += root->data;
 
        Leaves(root->right, sum_of_boundary_nodes);
    }
}
 
// Function to return the sum of all the
// boundary nodes of the given binary tree
int sumOfBoundaryNodes(struct Node* root)
{
    if (root) {
 
        // Root node is also a boundary node
        int sum_of_boundary_nodes = root->data;
 
        // Sum up all the left nodes
        // in TOP DOWN manner
        LeftBoundary(root->left, sum_of_boundary_nodes);
 
        // Sum up all the
        // leaf nodes
        Leaves(root->left, sum_of_boundary_nodes);
        Leaves(root->right, sum_of_boundary_nodes);
 
        // Sum up all the right nodes
        // in BOTTOM UP manner
        RightBoundary(root->right, sum_of_boundary_nodes);
 
        // Return the sum of
        // all the boundary nodes
        return sum_of_boundary_nodes;
    }
 
    return 0;
}
 
// Driver code
int main()
{
    Node* root = newNode(10);
    root->left = newNode(2);
    root->right = newNode(5);
    root->left->left = newNode(8);
    root->left->right = newNode(14);
    root->right->left = newNode(11);
    root->right->right = newNode(3);
    root->left->right->left = newNode(12);
    root->right->left->right = newNode(1);
    root->right->left->left = newNode(7);
 
    cout << sumOfBoundaryNodes(root);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    static int sum_of_boundary_nodes=0;
 
// A binary tree node has data,
// pointer to left child
static class Node
{
    int data;
    Node left;
    Node right;
};
 
// Utility function to create a node
static Node newNode(int data)
{
    Node temp = new Node();
 
    temp.left = null;
    temp.right = null;
    temp.data = data;
 
    return temp;
}
 
// Function to sum up all the left boundary nodes
// except the leaf nodes
static void LeftBoundary(Node root)
{
    if (root != null)
    {
        if (root.left != null)
        {
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.left);
        }
        else if (root.right != null)
        {
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.right);
        }
    }
}
 
// Function to sum up all the right boundary nodes
// except the leaf nodes
static void RightBoundary(Node root)
{
    if (root != null)
    {
        if (root.right != null)
        {
            RightBoundary(root.right);
            sum_of_boundary_nodes += root.data;
        }
        else if (root.left != null)
        {
            RightBoundary(root.left);
            sum_of_boundary_nodes += root.data;
        }
    }
}
 
// Function to sum up all the leaf nodes
// of a binary tree
static void Leaves(Node root)
{
    if (root != null)
    {
        Leaves(root.left);
 
        // Sum it up if it is a leaf node
        if ((root.left == null) && (root.right == null))
            sum_of_boundary_nodes += root.data;
 
        Leaves(root.right);
    }
}
 
// Function to return the sum of all the
// boundary nodes of the given binary tree
static int sumOfBoundaryNodes( Node root)
{
    if (root != null)
    {
 
        // Root node is also a boundary node
        sum_of_boundary_nodes = root.data;
 
        // Sum up all the left nodes
        // in TOP DOWN manner
        LeftBoundary(root.left);
 
        // Sum up all the
        // leaf nodes
        Leaves(root.left);
        Leaves(root.right);
 
        // Sum up all the right nodes
        // in BOTTOM UP manner
        RightBoundary(root.right);
 
        // Return the sum of
        // all the boundary nodes
        return sum_of_boundary_nodes;
    }
 
    return 0;
}
 
// Driver code
public static void main(String args[])
{
    Node root = newNode(10);
    root.left = newNode(2);
    root.right = newNode(5);
    root.left.left = newNode(8);
    root.left.right = newNode(14);
    root.right.left = newNode(11);
    root.right.right = newNode(3);
    root.left.right.left = newNode(12);
    root.right.left.right = newNode(1);
    root.right.left.left = newNode(7);
 
    System.out.println(sumOfBoundaryNodes(root));
}
}
 
// This code is contributed by andrew1234

Python3

# Python3 implementation of the approach
  
# A binary tree node has data,
# pointer to left child
# and a pointer to right child
class Node:
     
    def __init__(self):
         
        self.left = None
        self.right = None
         
sum_of_boundary_nodes = 0
 
# Utility function to create a node
def newNode(data):
 
    temp = Node()
    temp.data = data;
    return temp;
 
# Function to sum up all the
# left boundary nodes except
# the leaf nodes
def LeftBoundary(root):
     
    global sum_of_boundary_nodes
     
    if (root != None):
        if (root.left != None):
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.left);
         
        elif (root.right != None):
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.right);
 
# Function to sum up all the right
# boundary nodes except the leaf nodes
def RightBoundary(root):
     
    global sum_of_boundary_nodes
     
    if (root != None):
        if (root.right != None):
            RightBoundary(root.right);
            sum_of_boundary_nodes += root.data;
         
        elif (root.left != None):
            RightBoundary(root.left);
            sum_of_boundary_nodes += root.data;
         
# Function to sum up all the leaf nodes
# of a binary tree
def Leaves(root):
     
    global sum_of_boundary_nodes
     
    if (root != None):
        Leaves(root.left);
  
        # Sum it up if it is a leaf node
        if ((root.left == None) and
           (root.right == None)):
            sum_of_boundary_nodes += root.data;
  
        Leaves(root.right);
     
# Function to return the sum of all the
# boundary nodes of the given binary tree
def sumOfBoundaryNodes(root):
     
    global sum_of_boundary_nodes
     
    if (root != None):
         
        # Root node is also a boundary node
        sum_of_boundary_nodes = root.data;
  
        # Sum up all the left nodes
        # in TOP DOWN manner
        LeftBoundary(root.left);
  
        # Sum up all the
        # leaf nodes
        Leaves(root.left);
        Leaves(root.right);
  
        # Sum up all the right nodes
        # in BOTTOM UP manner
        RightBoundary(root.right);
  
        # Return the sum of
        # all the boundary nodes
        return sum_of_boundary_nodes;
     
    return 0;
 
# Driver code
if __name__=="__main__":
     
    root = newNode(10);
    root.left = newNode(2);
    root.right = newNode(5);
    root.left.left = newNode(8);
    root.left.right = newNode(14);
    root.right.left = newNode(11);
    root.right.right = newNode(3);
    root.left.right.left = newNode(12);
    root.right.left.right = newNode(1);
    root.right.left.left = newNode(7);
  
    print(sumOfBoundaryNodes(root));
 
# This code is contributed by rutvik_56

C#

// C# implementation of the approach
using System;
class GFG
{
static int sum_of_boundary_nodes = 0;
 
// A binary tree node has data,
// pointer to left child
public class Node
{
    public int data;
    public Node left;
    public Node right;
};
 
// Utility function to create a node
static Node newNode(int data)
{
    Node temp = new Node();
 
    temp.left = null;
    temp.right = null;
    temp.data = data;
 
    return temp;
}
 
// Function to sum up all the left boundary
// nodes except the leaf nodes
static void LeftBoundary(Node root)
{
    if (root != null)
    {
        if (root.left != null)
        {
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.left);
        }
         
        else if (root.right != null)
        {
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.right);
        }
    }
}
 
// Function to sum up all the right boundary
// nodes except the leaf nodes
static void RightBoundary(Node root)
{
    if (root != null)
    {
        if (root.right != null)
        {
            RightBoundary(root.right);
            sum_of_boundary_nodes += root.data;
        }
        else if (root.left != null)
        {
            RightBoundary(root.left);
            sum_of_boundary_nodes += root.data;
        }
    }
}
 
// Function to sum up all the leaf nodes
// of a binary tree
static void Leaves(Node root)
{
    if (root != null)
    {
        Leaves(root.left);
 
        // Sum it up if it is a leaf node
        if ((root.left == null) &&
            (root.right == null))
            sum_of_boundary_nodes += root.data;
 
        Leaves(root.right);
    }
}
 
// Function to return the sum of all the
// boundary nodes of the given binary tree
static int sumOfBoundaryNodes(Node root)
{
    if (root != null)
    {
 
        // Root node is also a boundary node
        sum_of_boundary_nodes = root.data;
 
        // Sum up all the left nodes
        // in TOP DOWN manner
        LeftBoundary(root.left);
 
        // Sum up all the
        // leaf nodes
        Leaves(root.left);
        Leaves(root.right);
 
        // Sum up all the right nodes
        // in BOTTOM UP manner
        RightBoundary(root.right);
 
        // Return the sum of
        // all the boundary nodes
        return sum_of_boundary_nodes;
    }
    return 0;
}
 
// Driver code
public static void Main(String []args)
{
    Node root = newNode(10);
    root.left = newNode(2);
    root.right = newNode(5);
    root.left.left = newNode(8);
    root.left.right = newNode(14);
    root.right.left = newNode(11);
    root.right.right = newNode(3);
    root.left.right.left = newNode(12);
    root.right.left.right = newNode(1);
    root.right.left.left = newNode(7);
 
    Console.WriteLine(sumOfBoundaryNodes(root));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>
 
    // JavaScript implementation of the approach
     
    let sum_of_boundary_nodes=0;
     
    // Binary Tree Node
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
 
    // Utility function to create a node
    function newNode(data)
    {
        let temp = new Node(data);
        return temp;
    }
 
    // Function to sum up all the left boundary nodes
    // except the leaf nodes
    function LeftBoundary(root)
    {
        if (root != null)
        {
            if (root.left != null)
            {
                sum_of_boundary_nodes += root.data;
                LeftBoundary(root.left);
            }
            else if (root.right != null)
            {
                sum_of_boundary_nodes += root.data;
                LeftBoundary(root.right);
            }
        }
    }
 
    // Function to sum up all the right boundary nodes
    // except the leaf nodes
    function RightBoundary(root)
    {
        if (root != null)
        {
            if (root.right != null)
            {
                RightBoundary(root.right);
                sum_of_boundary_nodes += root.data;
            }
            else if (root.left != null)
            {
                RightBoundary(root.left);
                sum_of_boundary_nodes += root.data;
            }
        }
    }
 
    // Function to sum up all the leaf nodes
    // of a binary tree
    function Leaves(root)
    {
        if (root != null)
        {
            Leaves(root.left);
 
            // Sum it up if it is a leaf node
            if ((root.left == null) && (root.right == null))
                sum_of_boundary_nodes += root.data;
 
            Leaves(root.right);
        }
    }
 
    // Function to return the sum of all the
    // boundary nodes of the given binary tree
    function sumOfBoundaryNodes(root)
    {
        if (root != null)
        {
 
            // Root node is also a boundary node
            sum_of_boundary_nodes = root.data;
 
            // Sum up all the left nodes
            // in TOP DOWN manner
            LeftBoundary(root.left);
 
            // Sum up all the
            // leaf nodes
            Leaves(root.left);
            Leaves(root.right);
 
            // Sum up all the right nodes
            // in BOTTOM UP manner
            RightBoundary(root.right);
 
            // Return the sum of
            // all the boundary nodes
            return sum_of_boundary_nodes;
        }
 
        return 0;
    }
     
    let root = newNode(10);
    root.left = newNode(2);
    root.right = newNode(5);
    root.left.left = newNode(8);
    root.left.right = newNode(14);
    root.right.left = newNode(11);
    root.right.right = newNode(3);
    root.left.right.left = newNode(12);
    root.right.left.right = newNode(1);
    root.right.left.left = newNode(7);
  
    document.write(sumOfBoundaryNodes(root));
 
</script>
Producción: 

48

 

Complejidad de tiempo: O(N) donde N es el número de Nodes en el árbol binario.
 

Publicación traducida automáticamente

Artículo escrito por Sakshi_Srivastava y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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