Dado un árbol binario, ¿cómo cuenta todos los medios Nodes (que tiene solo un hijo) sin usar la recursividad? Las hojas de nota no deben tocarse ya que tienen ambos hijos como NULL.
Input : Root of below tree
C++
// C++ program to count half 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 half Nodes in
// a binary tree
unsigned int gethalfCount(struct Node* node)
{
// If tree is empty
if (!node)
return 0;
int count = 0; // Initialize count of half nodes
// Do level order traversal starting from root
queue<Node *> q;
q.push(node);
while (!q.empty())
{
struct Node *temp = q.front();
q.pop();
if (!temp->left && temp->right ||
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 shown in
above example */
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 << gethalfCount(root);
return 0;
}
Java
// Java program to count half 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 half nodes of Tree
class BinaryTree
{
Node root;
/* Function to get the count of half Nodes in
a binary tree*/
int gethalfCount()
{
// If tree is empty
if (root==null)
return 0;
// Do level order traversal starting from root
Queue<Node> queue = new LinkedList<Node>();
queue.add(root);
int count=0; // Initialize count of half nodes
while (!queue.isEmpty())
{
Node temp = queue.poll();
if (temp.left!=null && temp.right==null ||
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 shown in
above example */
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.gethalfCount());
}
}
Python3
# Python program to count # half 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 half nodes of binary tree def gethalfCount(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 half nodes while(len(queue) > 0): node = queue.pop(0) # if it is half node then increment count if node.left is not None and node.right is None or node.left is 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 "%d" %(gethalfCount(root))
C#
// C# program to count half 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 half nodes of Tree
public class BinaryTree
{
Node root;
/* Function to get the count of half Nodes in
a binary tree*/
int gethalfCount()
{
// If tree is empty
if (root == null)
return 0;
// Do level order traversal starting from root
Queue<Node> queue = new Queue<Node>();
queue.Enqueue(root);
int count = 0; // Initialize count of half nodes
while (queue.Count != 0)
{
Node temp = queue.Dequeue();
if (temp.left != null && temp.right == null ||
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()
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
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.gethalfCount());
}
}
// This code contributed by Rajput-Ji
Javascript
<script>
// Javascript program to count half 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;
}
}
// Class to count half nodes of Tree
var root;
/* Function to get the count of half Nodes in
a binary tree*/
function gethalfCount()
{
// If tree is empty
if (root == null)
return 0;
// Do level order traversal starting from root
var queue = [];
queue.push(root);
var count = 0; // Initialize count of half nodes
while (queue.length != 0)
{
var temp = queue.shift();
if (temp.left != null && temp.right == null ||
temp.left == null && temp.right != null)
count++;
// push left child
if (temp.left != null)
queue.push(temp.left);
// push 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 shown in
above example */
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(gethalfCount());
// This code is contributed by famously.
</script>
C++
// C++ program to count half 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 half Nodes in
// a binary tree
unsigned int gethalfCount(struct Node* root)
{
if (root == NULL)
return 0;
int res = 0;
if ((root->left == NULL && root->right != NULL) ||
(root->left != NULL && root->right == NULL))
res++;
res += (gethalfCount(root->left) + gethalfCount(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 shown in
above example */
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 << gethalfCount(root);
return 0;
}
Java
// Java program to count half 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 half Nodes in
// a binary tree
static int gethalfCount(Node root)
{
if (root == null)
return 0;
int res = 0;
if ((root.left == null && root.right != null) ||
(root.left != null && root.right == null))
res++;
res += (gethalfCount(root.left)
+ gethalfCount(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 shown in
above example */
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(gethalfCount(root));
}
}
Python3
# Python program to count half nodes in a binary tree # A node structure class newNode: def __init__(self, data): self.data = data self.left = self.right = None # Function to get the count of half Nodes in a binary tree def gethalfCount(root): if root == None: return 0 res = 0 if(root.left == None and root.right != None) or \ (root.left != None and root.right == None): res += 1 res += (gethalfCount(root.left) + \ gethalfCount(root.right)) return res # Driver program ''' 2 / \ 7 5 \ \ 6 9 / \ / 1 11 4 Let us create Binary Tree shown in above example ''' 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(gethalfCount(root)) # This code is contributed by simranjenny84
C#
// C# program to count half 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 half Nodes in
// a binary tree
static int gethalfCount(Node root)
{
if (root == null)
return 0;
int res = 0;
if ((root.left == null && root.right != null) ||
(root.left != null && root.right == null))
res++;
res += (gethalfCount(root.left)
+ gethalfCount(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 code
public static void Main()
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
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(gethalfCount(root));
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
<script>
// Javascript program to count half
// nodes in a Binary Tree
class Node
{
constructor(data)
{
this.left = null;
this.right = null;
this.data = data;
}
}
// Function to get the count of half
// Nodes in a binary tree
function gethalfCount(root)
{
if (root == null)
return 0;
let res = 0;
if ((root.left == null && root.right != null) ||
(root.left != null && root.right == null))
res++;
res += (gethalfCount(root.left) +
gethalfCount(root.right));
return res;
}
/* Helper function that allocates a new
Node with the given data and NULL left
and right pointers. */
function newNode(data)
{
let node = new Node(data);
return (node);
}
// Driver code
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
let 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(gethalfCount(root));
// This code is contributed by divyeshrabadiya07
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA