Dado un árbol binario, devuelve el siguiente valor para él.
1) Para cada nivel, calcule la suma de todas las hojas si hay hojas en este nivel. De lo contrario, ignóralo.
2) Devuelve la multiplicación de todas las sumas.
Ejemplos:
Input: Root of below tree
2
/ \
7 5
\
9
Output: 63
First levels doesn't have leaves. Second level
has one leaf 7 and third level also has one
leaf 9. Therefore result is 7*9 = 63
Input: Root of below tree
2
/ \
7 5
/ \ \
8 6 9
/ \ / \
1 11 4 10
Output: 208
First two levels don't have leaves. Third
level has single leaf 8. Last level has four
leaves 1, 11, 4 and 10. Therefore result is
8 * (1 + 11 + 4 + 10)
Le recomendamos encarecidamente que minimice su navegador y que pruebe esto usted mismo primero.
Una solución simple es calcular recursivamente la suma de hojas para todos los niveles, comenzando de arriba a abajo. Luego multiplica las sumas de los niveles que tienen hojas. La complejidad temporal de esta solución sería O(n 2 ).
Una solución eficiente es utilizar el cruce de orden de nivel basado en cola. Mientras realiza el recorrido, procese todos los niveles diferentes por separado. Para cada nivel procesado, verifique si tiene hojas. Si lo ha hecho, calcule la suma de los Nodes hoja. Finalmente devuelva el producto de todas las sumas.
C++
/* Iterative C++ program to find sum of data of all leaves
of a binary tree on same level and then multiply sums
obtained of all levels. */
#include <bits/stdc++.h>
using namespace std;
// A Binary Tree Node
struct Node {
int data;
struct Node *left, *right;
};
// helper function to check if a Node is leaf of tree
bool isLeaf(Node* root)
{
return (!root->left && !root->right);
}
/* Calculate sum of all leaf Nodes at each level and returns
multiplication of sums */
int sumAndMultiplyLevelData(Node* root)
{
// Tree is empty
if (!root)
return 0;
int mul = 1; /* To store result */
// Create an empty queue for level order traversal
queue<Node*> q;
// Enqueue Root and initialize height
q.push(root);
// Do level order traversal of tree
while (1) {
// NodeCount (queue size) indicates number of Nodes
// at current level.
int NodeCount = q.size();
// If there are no Nodes at current level, we are done
if (NodeCount == 0)
break;
// Initialize leaf sum for current level
int levelSum = 0;
// A boolean variable to indicate if found a leaf
// Node at current level or not
bool leafFound = false;
// Dequeue all Nodes of current level and Enqueue all
// Nodes of next level
while (NodeCount > 0) {
// Process next Node of current level
Node* Node = q.front();
/* if Node is a leaf, update sum at the level */
if (isLeaf(Node)) {
leafFound = true;
levelSum += Node->data;
}
q.pop();
// Add children of Node
if (Node->left != NULL)
q.push(Node->left);
if (Node->right != NULL)
q.push(Node->right);
NodeCount--;
}
// If we found at least one leaf, we multiply
// result with level sum.
if (leafFound)
mul *= levelSum;
}
return mul; // Return result
}
// Utility function to create a new tree Node
Node* newNode(int data)
{
Node* temp = new Node;
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
// Driver program to test above functions
int main()
{
Node* root = newNode(2);
root->left = newNode(7);
root->right = newNode(5);
root->left->right = newNode(6);
root->left->left = newNode(8);
root->left->right->left = newNode(1);
root->left->right->right = newNode(11);
root->right->right = newNode(9);
root->right->right->left = newNode(4);
root->right->right->right = newNode(10);
cout << "Final product value = "
<< sumAndMultiplyLevelData(root) << endl;
return 0;
}
Java
/* Iterative Java program to find sum of data of all leaves
of a binary tree on same level and then multiply sums
obtained of all levels. */
/* importing the necessary class */
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
/* Class containing left and right child of current
node and key value*/
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
class BinaryTree {
Node root;
// helper function to check if a Node is leaf of tree
boolean isLeaf(Node node)
{
return ((node.left == null) && (node.right == null));
}
/* Calculate sum of all leaf Nodes at each level and returns
multiplication of sums */
int sumAndMultiplyLevelData()
{
return sumAndMultiplyLevelData(root);
}
int sumAndMultiplyLevelData(Node node)
{
// Tree is empty
if (node == null) {
return 0;
}
int mul = 1; /* To store result */
// Create an empty queue for level order traversal
LinkedList<Node> q = new LinkedList<Node>();
// Enqueue Root and initialize height
q.add(node);
// Do level order traversal of tree
while (true) {
// NodeCount (queue size) indicates number of Nodes
// at current level.
int NodeCount = q.size();
// If there are no Nodes at current level, we are done
if (NodeCount == 0) {
break;
}
// Initialize leaf sum for current level
int levelSum = 0;
// A boolean variable to indicate if found a leaf
// Node at current level or not
boolean leafFound = false;
// Dequeue all Nodes of current level and Enqueue all
// Nodes of next level
while (NodeCount > 0) {
Node node1;
node1 = q.poll();
/* if Node is a leaf, update sum at the level */
if (isLeaf(node1)) {
leafFound = true;
levelSum += node1.data;
}
// Add children of Node
if (node1.left != null) {
q.add(node1.left);
}
if (node1.right != null) {
q.add(node1.right);
}
NodeCount--;
}
// If we found at least one leaf, we multiply
// result with level sum.
if (leafFound) {
mul *= levelSum;
}
}
return mul; // Return result
}
public static void main(String args[])
{
/* creating a binary tree and entering
the nodes */
BinaryTree tree = new BinaryTree();
tree.root = new Node(2);
tree.root.left = new Node(7);
tree.root.right = new Node(5);
tree.root.left.left = new Node(8);
tree.root.left.right = new Node(6);
tree.root.left.right.left = new Node(1);
tree.root.left.right.right = new Node(11);
tree.root.right.right = new Node(9);
tree.root.right.right.left = new Node(4);
tree.root.right.right.right = new Node(10);
System.out.println("The final product value : "
+ tree.sumAndMultiplyLevelData());
}
}
// This code is contributed by Mayank Jaiswal
Python3
"""Iterative Python3 program to find
sum of data of all leaves of a binary
tree on same level and then multiply
sums obtained of all levels."""
# A Binary Tree Node
# Utility function to create a
# new tree Node
class newNode:
def __init__(self, data):
self.data = data
self.left = self.right = None
# helper function to check if a
# Node is leaf of tree
def isLeaf(root) :
return (not root.left and
not root.right)
""" Calculate sum of all leaf Nodes at each
level and returns multiplication of sums """
def sumAndMultiplyLevelData( root) :
# Tree is empty
if (not root) :
return 0
mul = 1
""" To store result """
# Create an empty queue for level
# order traversal
q = []
# Enqueue Root and initialize height
q.append(root)
# Do level order traversal of tree
while (1):
# NodeCount (queue size) indicates
# number of Nodes at current level.
NodeCount = len(q)
# If there are no Nodes at current
# level, we are done
if (NodeCount == 0) :
break
# Initialize leaf sum for
# current level
levelSum = 0
# A boolean variable to indicate
# if found a leaf Node at current
# level or not
leafFound = False
# Dequeue all Nodes of current level
# and Enqueue all Nodes of next level
while (NodeCount > 0) :
# Process next Node of current level
Node = q[0]
""" if Node is a leaf, update
sum at the level """
if (isLeaf(Node)) :
leafFound = True
levelSum += Node.data
q.pop(0)
# Add children of Node
if (Node.left != None) :
q.append(Node.left)
if (Node.right != None) :
q.append(Node.right)
NodeCount-=1
# If we found at least one leaf,
# we multiply result with level sum.
if (leafFound) :
mul *= levelSum
return mul # Return result
# Driver Code
if __name__ == '__main__':
root = newNode(2)
root.left = newNode(7)
root.right = newNode(5)
root.left.right = newNode(6)
root.left.left = newNode(8)
root.left.right.left = newNode(1)
root.left.right.right = newNode(11)
root.right.right = newNode(9)
root.right.right.left = newNode(4)
root.right.right.right = newNode(10)
print("Final product value = ",
sumAndMultiplyLevelData(root))
# This code is contributed
# by SHUBHAMSINGH10
C#
/* Iterative C# program to find sum
of data of all leaves of a binary tree
on same level and then multiply sums
obtained of all levels. */
/* importing the necessary class */
using System;
using System.Collections.Generic;
/* Class containing left and right child
of current node and key value*/
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
public class BinaryTree
{
Node root;
// helper function to check if
// a Node is leaf of tree
bool isLeaf(Node node)
{
return ((node.left == null) &&
(node.right == null));
}
/* Calculate sum of all leaf
Nodes at each level and returns
multiplication of sums */
int sumAndMultiplyLevelData()
{
return sumAndMultiplyLevelData(root);
}
int sumAndMultiplyLevelData(Node node)
{
// Tree is empty
if (node == null) {
return 0;
}
int mul = 1; /* To store result */
// Create an empty queue for level order traversal
Queue<Node> q = new Queue<Node>();
// Enqueue Root and initialize height
q.Enqueue(node);
// Do level order traversal of tree
while (true) {
// NodeCount (queue size) indicates
// number of Nodes at current level.
int NodeCount = q.Count;
// If there are no Nodes at current
// level, we are done
if (NodeCount == 0)
{
break;
}
// Initialize leaf sum for current level
int levelSum = 0;
// A boolean variable to indicate if found a leaf
// Node at current level or not
bool leafFound = false;
// Dequeue all Nodes of current level and
// Enqueue all Nodes of next level
while (NodeCount > 0)
{
Node node1;
node1 = q.Dequeue();
/* if Node is a leaf, update sum at the level */
if (isLeaf(node1))
{
leafFound = true;
levelSum += node1.data;
}
// Add children of Node
if (node1.left != null)
{
q.Enqueue(node1.left);
}
if (node1.right != null)
{
q.Enqueue(node1.right);
}
NodeCount--;
}
// If we found at least one leaf, we multiply
// result with level sum.
if (leafFound)
{
mul *= levelSum;
}
}
return mul; // Return result
}
// Driver code
public static void Main(String []args)
{
/* creating a binary tree and entering
the nodes */
BinaryTree tree = new BinaryTree();
tree.root = new Node(2);
tree.root.left = new Node(7);
tree.root.right = new Node(5);
tree.root.left.left = new Node(8);
tree.root.left.right = new Node(6);
tree.root.left.right.left = new Node(1);
tree.root.left.right.right = new Node(11);
tree.root.right.right = new Node(9);
tree.root.right.right.left = new Node(4);
tree.root.right.right.right = new Node(10);
Console.WriteLine("The final product value : "
+ tree.sumAndMultiplyLevelData());
}
}
// This code has been contributed by 29AjayKumar
Javascript
<script>
/* Iterative Javascript program to
find sum of data of all leaves
of a binary tree on same level
and then multiply sums
obtained of all levels. */
/* importing the necessary class */
/* Class containing left and right child of current
node and key value*/
class Node
{
constructor(data)
{
this.data=data;
this.left = this.right = null;
}
}
let root;
// helper function to check if a Node is leaf of tree
function isLeaf(node)
{
return ((node.left == null) && (node.right == null));
}
/* Calculate sum of all leaf Nodes at each level and returns
multiplication of sums */
function sumAndMultiplyLevelData(node)
{
// Tree is empty
if (node == null) {
return 0;
}
let mul = 1; /* To store result */
// Create an empty queue for level order traversal
let q = [];
// Enqueue Root and initialize height
q.push(node);
// Do level order traversal of tree
while (true) {
// NodeCount (queue size) indicates number of Nodes
// at current level.
let NodeCount = q.length;
// If there are no Nodes at current level, we are done
if (NodeCount == 0) {
break;
}
// Initialize leaf sum for current level
let levelSum = 0;
// A boolean variable to indicate if found a leaf
// Node at current level or not
let leafFound = false;
// Dequeue all Nodes of current level and Enqueue all
// Nodes of next level
while (NodeCount > 0) {
let node1= q.shift();
/* if Node is a leaf, update sum at the level */
if (isLeaf(node1)) {
leafFound = true;
levelSum += node1.data;
}
// Add children of Node
if (node1.left != null) {
q.push(node1.left);
}
if (node1.right != null) {
q.push(node1.right);
}
NodeCount--;
}
// If we found at least one leaf, we multiply
// result with level sum.
if (leafFound) {
mul *= levelSum;
}
}
return mul; // Return result
}
/* creating a binary tree and entering
the nodes */
root = new Node(2);
root.left = new Node(7);
root.right = new Node(5);
root.left.left = new Node(8);
root.left.right = new Node(6);
root.left.right.left = new Node(1);
root.left.right.right = new Node(11);
root.right.right = new Node(9);
root.right.right.left = new Node(4);
root.right.right.right = new Node(10);
document.write("The final product value : "
+ sumAndMultiplyLevelData(root));
// This code is contributed by rag2127
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA