Dado un árbol binario, imprima los Nodes de las esquinas en cada nivel. El Node de más a la izquierda y el Node de más a la derecha.
Por ejemplo, la salida para seguir es 15, 10, 20, 8, 25 .
C++
// C/C++ program to print corner node at each level
// of binary tree
#include <bits/stdc++.h>
using namespace std;
/* A binary tree node has key, pointer to left
child and a pointer to right child */
struct Node
{
int key;
struct Node* left, *right;
};
/* To create a newNode of tree and return pointer */
struct Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return (temp);
}
/* Function to print corner node at each level */
void printCorner(Node *root)
{
//If the root is null then simply return
if(root == NULL)
return;
//Do level order traversal using a single queue
queue<Node*> q;
q.push(root);
while(!q.empty())
{
//n denotes the size of the current level in the queue
int n = q.size();
for(int i =0;i<n;i++)
{
Node *temp = q.front();
q.pop();
//If it is leftmost corner value or rightmost corner value then print it
if(i==0 || i==n-1)
cout<<temp->key<<" ";
//push the left and right children of the temp node
if(temp->left)
q.push(temp->left);
if(temp->right)
q.push(temp->right);
}
}
}
// Driver program to test above function
int main ()
{
Node *root = newNode(15);
root->left = newNode(10);
root->right = newNode(20);
root->left->left = newNode(8);
root->left->right = newNode(12);
root->right->left = newNode(16);
root->right->right = newNode(25);
printCorner(root);
return 0;
}
// This code is contributed by Tapeshdua420.
Java
// Java program to print corner node at each level in a binary tree
import java.util.*;
/* A binary tree node has key, pointer to left
child and a pointer to right child */
class Node
{
int key;
Node left, right;
public Node(int key)
{
this.key = key;
left = right = null;
}
}
class BinaryTree
{
Node root;
/* Function to print corner node at each level */
void printCorner(Node root)
{
// star node is for keeping track of levels
Queue<Node> q = new LinkedList<Node>();
// pushing root node and star node
q.add(root);
// Do level order traversal of Binary Tree
while (!q.isEmpty())
{
// n is the no of nodes in current Level
int n = q.size();
for(int i = 0 ; i < n ; i++){
// dequeue the front node from the queue
Node temp = q.peek();
q.poll();
//If it is leftmost corner value or rightmost corner value then print it
if(i==0 || i==n-1)
System.out.print(temp.key + " ");
//push the left and right children of the temp node
if (temp.left != null)
q.add(temp.left);
if (temp.right != null)
q.add(temp.right);
}
}
}
// Driver program to test above functions
public static void main(String[] args)
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(15);
tree.root.left = new Node(10);
tree.root.right = new Node(20);
tree.root.left.left = new Node(8);
tree.root.left.right = new Node(12);
tree.root.right.left = new Node(16);
tree.root.right.right = new Node(25);
tree.printCorner(tree.root);
}
}
// This code has been contributed by Utkarsh Choubey
Python3
# Python3 program to print corner # node at each level of binary tree from collections import deque # A binary tree node has key, pointer to left # child and a pointer to right child class Node: def __init__(self, key): self.key = key self.left = None self.right = None # Function to print corner node at each level def printCorner(root: Node): # If the root is null then simply return if root == None: return # Do level order traversal # using a single queue q = deque() q.append(root) while q: # n denotes the size of the current # level in the queue n = len(q) for i in range(n): temp = q[0] q.popleft() # If it is leftmost corner value or # rightmost corner value then print it if i == 0 or i == n - 1: print(temp.key, end = " ") # push the left and right children # of the temp node if temp.left: q.append(temp.left) if temp.right: q.append(temp.right) # Driver Code if __name__ == "__main__": root = Node(15) root.left = Node(10) root.right = Node(20) root.left.left = Node(8) root.left.right = Node(12) root.right.left = Node(16) root.right.right = Node(25) printCorner(root) # This code is contributed by sanjeev2552
C#
// C# program to print corner node
// at each level in a binary tree
using System;
using System.Collections.Generic;
/* A binary tree node has key, pointer to left
child and a pointer to right child */
public class Node
{
public int key;
public Node left, right;
public Node(int key)
{
this.key = key;
left = right = null;
}
}
public class BinaryTree
{
Node root;
/* Function to print corner node at each level */
void printCorner(Node root)
{
// star node is for keeping track of levels
Queue<Node> q = new Queue<Node>();
// pushing root node and star node
q.Enqueue(root);
// Do level order traversal of Binary Tree
while (q.Count != 0)
{
// n is the no of nodes in current Level
int n = q.Count;
for(int i = 0 ; i < n ; i++){
Node temp = q.Peek();
q.Dequeue();
//If it is leftmost corner value or rightmost corner value then print it
if(i==0||i==n-1)
Console.Write(temp.key + " ");
//push the left and right children of the temp node
if (temp.left != null)
q.Enqueue(temp.left);
if (temp.right != null)
q.Enqueue(temp.right);
}
}
}
// Driver code
public static void Main(String[] args)
{
BinaryTree tree = new BinaryTree();
tree.root = new Node(15);
tree.root.left = new Node(10);
tree.root.right = new Node(20);
tree.root.left.left = new Node(8);
tree.root.left.right = new Node(12);
tree.root.right.left = new Node(16);
tree.root.right.right = new Node(25);
tree.printCorner(tree.root);
}
}
// This code is contributed by Utkarsh Choubey
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA