Dado un árbol binario, ¿cómo cuenta todos los Nodes completos (Nodes que tienen tanto hijos como no NULL) sin usar la recursividad y con la recursividad? Las hojas de nota no deben tocarse ya que tienen ambos hijos como NULL.
Los Nodes 2 y 6 son Nodes completos y tienen ambos hijos. Entonces, el recuento de Nodes completos en el árbol anterior es 2
Método: Iterativo
La idea es utilizar el recorrido de orden de nivel para resolver este problema de manera eficiente.
1) Create an empty Queue Node and push root node to Queue.
2) Do following while nodeQeue is not empty.
a) Pop an item from Queue and process it.
a.1) If it is full node then increment count++.
b) Push left child of popped item to Queue, if available.
c) Push right child of popped item to Queue, if available.
C++
// C++ program to count
// full nodes in a Binary Tree
#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;
};
// Function to get the count of full Nodes in
// a binary tree
unsigned int getfullCount(struct Node* node)
{
// If tree is empty
if (!node)
return 0;
queue<Node *> q;
// Do level order traversal starting from root
int count = 0; // Initialize count of full nodes
q.push(node);
while (!q.empty())
{
struct Node *temp = q.front();
q.pop();
if (temp->left && temp->right)
count++;
if (temp->left != NULL)
q.push(temp->left);
if (temp->right != NULL)
q.push(temp->right);
}
return count;
}
/* 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 = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Driver program
int main(void)
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree as shown
*/
struct Node *root = newNode(2);
root->left = newNode(7);
root->right = newNode(5);
root->left->right = newNode(6);
root->left->right->left = newNode(1);
root->left->right->right = newNode(11);
root->right->right = newNode(9);
root->right->right->left = newNode(4);
cout << getfullCount(root);
return 0;
}
Java
// Java program to count
// full nodes in a Binary Tree
// using Iterative approach
import java.util.Queue;
import java.util.LinkedList;
// Class to represent Tree node
class Node
{
int data;
Node left, right;
public Node(int item)
{
data = item;
left = null;
right = null;
}
}
// Class to count full nodes of Tree
class BinaryTree
{
Node root;
/* Function to get the count of full Nodes in
a binary tree*/
int getfullCount()
{
// If tree is empty
if (root==null)
return 0;
// Initialize empty queue.
Queue<Node> queue = new LinkedList<Node>();
// Do level order traversal starting from root
queue.add(root);
int count=0; // Initialize count of full nodes
while (!queue.isEmpty())
{
Node temp = queue.poll();
if (temp.left!=null && temp.right!=null)
count++;
// Enqueue left child
if (temp.left != null)
{
queue.add(temp.left);
}
// Enqueue right child
if (temp.right != null)
{
queue.add(temp.right);
}
}
return count;
}
public static void main(String args[])
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree as shown
*/
BinaryTree tree_level = new BinaryTree();
tree_level.root = new Node(2);
tree_level.root.left = new Node(7);
tree_level.root.right = new Node(5);
tree_level.root.left.right = new Node(6);
tree_level.root.left.right.left = new Node(1);
tree_level.root.left.right.right = new Node(11);
tree_level.root.right.right = new Node(9);
tree_level.root.right.right.left = new Node(4);
System.out.println(tree_level.getfullCount());
}
}
Python3
# Python program to count # full nodes in a Binary Tree # using iterative approach # A node structure class Node: # A utility function to create a new node def __init__(self ,key): self.data = key self.left = None self.right = None # Iterative Method to count full nodes of binary tree def getfullCount(root): # Base Case if root is None: return 0 # Create an empty queue for level order traversal queue = [] # Enqueue Root and initialize count queue.append(root) count = 0 #initialize count for full nodes while(len(queue) > 0): node = queue.pop(0) # if it is full node then increment count if node.left is not None and node.right is not None: count = count+1 # Enqueue left child if node.left is not None: queue.append(node.left) # Enqueue right child if node.right is not None: queue.append(node.right) return count # Driver Program to test above function root = Node(2) root.left = Node(7) root.right = Node(5) root.left.right = Node(6) root.left.right.left = Node(1) root.left.right.right = Node(11) root.right.right = Node(9) root.right.right.left = Node(4) print(getfullCount(root))
C#
// C# program to count
// full nodes in a Binary Tree
// using Iterative approach
using System;
using System.Collections.Generic;
// Class to represent Tree node
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = null;
right = null;
}
}
// Class to count full nodes of Tree
public class BinaryTree
{
Node root;
/* Function to get the count of full Nodes in
a binary tree*/
int getfullCount()
{
// If tree is empty
if (root == null)
return 0;
// Initialize empty queue.
Queue<Node> queue = new Queue<Node>();
// Do level order traversal starting from root
queue.Enqueue(root);
int count = 0; // Initialize count of full nodes
while (queue.Count != 0)
{
Node temp = queue.Dequeue();
if (temp.left != null && temp.right != null)
count++;
// Enqueue left child
if (temp.left != null)
{
queue.Enqueue(temp.left);
}
// Enqueue right child
if (temp.right != null)
{
queue.Enqueue(temp.right);
}
}
return count;
}
// Driver code
public static void Main(String []args)
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree as shown
*/
BinaryTree tree_level = new BinaryTree();
tree_level.root = new Node(2);
tree_level.root.left = new Node(7);
tree_level.root.right = new Node(5);
tree_level.root.left.right = new Node(6);
tree_level.root.left.right.left = new Node(1);
tree_level.root.left.right.right = new Node(11);
tree_level.root.right.right = new Node(9);
tree_level.root.right.right.left = new Node(4);
Console.WriteLine(tree_level.getfullCount());
}
}
// This code has been contributed by 29AjayKumar
Javascript
<script>
// Javascript program to count
// full nodes in a Binary Tree
// using Iterative approach
// Class to represent Tree node
class Node
{
constructor(item)
{
this.data = item;
this.left = null;
this.right = null;
}
}
let root;
// Function to get the count of full
// Nodes in a binary tree
function getfullCount()
{
// If tree is empty
if (root == null)
return 0;
// Initialize empty queue.
let queue = [];
// Do level order traversal starting from root
queue.push(root);
// Initialize count of full nodes
let count = 0;
while (queue.length != 0)
{
let temp = queue.shift();
if (temp.left != null && temp.right != null)
count++;
// Enqueue left child
if (temp.left != null)
{
queue.push(temp.left);
}
// Enqueue right child
if (temp.right != null)
{
queue.push(temp.right);
}
}
return count;
}
// Driver code
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree as shown
*/
root = new Node(2);
root.left = new Node(7);
root.right = new Node(5);
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);
document.write(getfullCount());
// This code is contributed by rag2127
</script>
C++
// C++ program to count full nodes in a Binary Tree
#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;
};
// Function to get the count of full Nodes in
// a binary tree
unsigned int getfullCount(struct Node* root)
{
if (root == NULL)
return 0;
int res = 0;
if (root->left && root->right)
res++;
res += (getfullCount(root->left) +
getfullCount(root->right));
return res;
}
/* 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 = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Driver program
int main(void)
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree as shown
*/
struct Node *root = newNode(2);
root->left = newNode(7);
root->right = newNode(5);
root->left->right = newNode(6);
root->left->right->left = newNode(1);
root->left->right->right = newNode(11);
root->right->right = newNode(9);
root->right->right->left = newNode(4);
cout << getfullCount(root);
return 0;
}
Java
// Java program to count full nodes in a Binary Tree
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;
}
// Function to get the count of full Nodes in
// a binary tree
static int getfullCount(Node root)
{
if (root == null)
return 0;
int res = 0;
if (root.left != null && root.right != null)
res++;
res += (getfullCount(root.left) + getfullCount(root.right));
return res;
}
/* Helper function that allocates a new
Node with the given data and NULL left
and right pointers. */
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Driver program
public static void main(String[] args)
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree as shown
*/
Node root = newNode(2);
root.left = newNode(7);
root.right = newNode(5);
root.left.right = newNode(6);
root.left.right.left = newNode(1);
root.left.right.right = newNode(11);
root.right.right = newNode(9);
root.right.right.left = newNode(4);
System.out.println(getfullCount(root));
}
}
Python3
# Python program to count full # nodes in a Binary Tree class newNode(): def __init__(self, data): self.data = data self.left = None self.right = None # Function to get the count of # full Nodes in a binary tree def getfullCount(root): if (root == None): return 0 res = 0 if (root.left and root.right): res += 1 res += (getfullCount(root.left) + getfullCount(root.right)) return res # Driver code if __name__ == '__main__': """ 2 / \ 7 5 \ \ 6 9 / \ / 1 11 4 Let us create Binary Tree as shown """ root = newNode(2) root.left = newNode(7) root.right = newNode(5) root.left.right = newNode(6) root.left.right.left = newNode(1) root.left.right.right = newNode(11) root.right.right = newNode(9) root.right.right.left = newNode(4) print(getfullCount(root)) # This code is contributed by SHUBHAMSINGH10
C#
// C# program to count full nodes in a Binary Tree
using System;
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;
}
// Function to get the count of full Nodes in
// a binary tree
static int getfullCount(Node root)
{
if (root == null)
return 0;
int res = 0;
if (root.left != null && root.right != null)
res++;
res += (getfullCount(root.left) + getfullCount(root.right));
return res;
}
/* Helper function that allocates a new
Node with the given data and NULL left
and right pointers. */
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Driver program
public static void Main()
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree as shown
*/
Node root = newNode(2);
root.left = newNode(7);
root.right = newNode(5);
root.left.right = newNode(6);
root.left.right.left = newNode(1);
root.left.right.right = newNode(11);
root.right.right = newNode(9);
root.right.right.left = newNode(4);
Console.WriteLine(getfullCount(root));
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
<script>
// JavaScript program to count full nodes in a Binary Tree
// A binary tree Node has data, pointer to left
// child and a pointer to right child
class Node
{
constructor()
{
this.data = 0;
this.left = null;
this.right = null;
}
}
// Function to get the count of full Nodes in
// a binary tree
function getfullCount(root)
{
if (root == null)
return 0;
var res = 0;
if (root.left != null && root.right != null)
res++;
res += (getfullCount(root.left) + getfullCount(root.right));
return res;
}
/* Helper function that allocates a new
Node with the given data and NULL left
and right pointers. */
function newNode(data)
{
var node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Driver program
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree as shown
*/
var root = newNode(2);
root.left = newNode(7);
root.right = newNode(5);
root.left.right = newNode(6);
root.left.right.left = newNode(1);
root.left.right.right = newNode(11);
root.right.right = newNode(9);
root.right.right.left = newNode(4);
document.write(getfullCount(root));
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA