Dado un árbol binario y un valor K . La tarea es imprimir el k-ésimo Node en el recorrido diagonal del árbol binario. Si no existe tal Node, imprima -1.
Ejemplos:
Input :
8
/ \
3 10
/ / \
1 6 14
/ \ /
4 7 13
k = 5
Output : 6
Diagonal Traversal of the above tree is:
8 10 14
3 6 7 13
1 4
Input :
1
/ \
2 3
/ \
4 5
k = 7
Output : -1
Enfoque: La idea es realizar el recorrido diagonal del árbol binario hasta que se visiten K Nodes en el recorrido diagonal. Mientras recorre cada Node visitado, disminuya el valor de la variable K y devuelva el Node actual cuando el valor de K sea cero. Si el recorrido diagonal no contiene al menos K Nodes, devuelve -1.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to print kth node
// in the diagonal traversal of 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;
Node *left, *right;
};
// Helper function that allocates a new node
Node* newNode(int data)
{
Node* node = new Node();
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Iterative function to print kth node
// in diagonal traversal of binary tree
int diagonalPrint(Node* root, int k)
{
// Base cases
if (root == NULL || k == 0)
return -1;
int ans = -1;
queue<Node*> q;
// Push root node
q.push(root);
// Push delimiter NULL
q.push(NULL);
while (!q.empty()) {
Node* temp = q.front();
q.pop();
if (temp == NULL) {
if (q.empty()) {
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break;
}
q.push(NULL);
}
else {
while (temp) {
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp->data;
if (temp->left)
q.push(temp->left);
temp = temp->right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
}
// Driver Code
int main()
{
Node* root = newNode(8);
root->left = newNode(3);
root->right = newNode(10);
root->left->left = newNode(1);
root->left->right = newNode(6);
root->right->right = newNode(14);
root->right->right->left = newNode(13);
root->left->right->left = newNode(4);
root->left->right->right = newNode(7);
int k = 9;
cout << diagonalPrint(root, k);
return 0;
}
Java
// Java program to print kth node
// in the diagonal traversal of 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;
};
// Helper function that allocates a new node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// Iterative function to print kth node
// in diagonal traversal of binary tree
static int diagonalPrint(Node root, int k)
{
// Base cases
if (root == null || k == 0)
return -1;
int ans = -1;
Queue<Node> q = new LinkedList<Node>();
// add root node
q.add(root);
// add delimiter null
q.add(null);
while (q.size() > 0)
{
Node temp = q.peek();
q.remove();
if (temp == null)
{
if (q.size() == 0)
{
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break;
}
q.add(null);
}
else {
while (temp != null)
{
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp.data;
if (temp.left!=null)
q.add(temp.left);
temp = temp.right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
}
// Driver Code
public static void main(String args[])
{
Node root = newNode(8);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(1);
root.left.right = newNode(6);
root.right.right = newNode(14);
root.right.right.left = newNode(13);
root.left.right.left = newNode(4);
root.left.right.right = newNode(7);
int k = 9;
System.out.println( diagonalPrint(root, k));
}
}
// This code is contributed by Arnab Kundu
Python3
# Python program to print kth node # in the diagonal traversal of a binary tree # Linked List node class Node: def __init__(self, data): self.data = data self.left = None self.right = None # Helper function that allocates a new node def newNode(data) : node = Node(0) node.data = data node.left = node.right = None return (node) # Iterative function to print kth node # in diagonal traversal of binary tree def diagonalPrint( root, k) : # Base cases if (root == None or k == 0) : return -1 ans = -1 q = [] # append root node q.append(root) # append delimiter None q.append(None) while (len(q) > 0): temp = q[0] q.pop(0) if (temp == None): if (len(q) == 0) : # If kth node exists then return # the answer if (k == 0) : return ans # If kth node doesnt exists # then break from the while loop else: break q.append(None) else : while (temp != None): # If the required kth node # has been found then return the answer if (k == 0) : return ans k = k - 1 # Update the value of variable ans # each time ans = temp.data if (temp.left != None): q.append(temp.left) temp = temp.right # If kth node doesnt exists then # return -1 return -1 # Driver Code root = newNode(8) root.left = newNode(3) root.right = newNode(10) root.left.left = newNode(1) root.left.right = newNode(6) root.right.right = newNode(14) root.right.right.left = newNode(13) root.left.right.left = newNode(4) root.left.right.right = newNode(7) k = 9 print( diagonalPrint(root, k)) # This code is contributed by Arnab Kundu
C#
// C# program to print kth node
// in the diagonal traversal of a binary tree
using System;
using System.Collections.Generic;
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;
};
// Helper function that allocates a new node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// Iterative function to print kth node
// in diagonal traversal of binary tree
static int diagonalPrint(Node root, int k)
{
// Base cases
if (root == null || k == 0)
return -1;
int ans = -1;
Queue<Node> q = new Queue<Node>();
// Enqueue root node
q.Enqueue(root);
// Enqueue delimiter null
q.Enqueue(null);
while (q.Count > 0)
{
Node temp = q.Peek();
q.Dequeue();
if (temp == null)
{
if (q.Count == 0)
{
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break;
}
q.Enqueue(null);
}
else
{
while (temp != null)
{
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp.data;
if (temp.left!=null)
q.Enqueue(temp.left);
temp = temp.right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
}
// Driver Code
public static void Main(String []args)
{
Node root = newNode(8);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(1);
root.left.right = newNode(6);
root.right.right = newNode(14);
root.right.right.left = newNode(13);
root.left.right.left = newNode(4);
root.left.right.right = newNode(7);
int k = 9;
Console.WriteLine( diagonalPrint(root, k));
}
}
/* This code is contributed by PrinciRaj1992 */
Javascript
<script>
// JavaScript program to print kth node
// in the diagonal traversal of 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;
}
};
// Helper function that allocates a new node
function newNode(data)
{
var node = new Node();
node.data = data;
node.left = node.right = null;
return (node);
}
// Iterative function to print kth node
// in diagonal traversal of binary tree
function diagonalPrint(root, k)
{
// Base cases
if (root == null || k == 0)
return -1;
var ans = -1;
var q = [];
// push root node
q.push(root);
// push delimiter null
q.push(null);
while (q.length > 0)
{
var temp = q[0];
q.shift();
if (temp == null)
{
if (q.length == 0)
{
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break;
}
q.push(null);
}
else
{
while (temp != null)
{
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp.data;
if (temp.left!=null)
q.push(temp.left);
temp = temp.right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
}
// Driver Code
var root = newNode(8);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(1);
root.left.right = newNode(6);
root.right.right = newNode(14);
root.right.right.left = newNode(13);
root.left.right.left = newNode(4);
root.left.right.right = newNode(7);
var k = 9;
document.write( diagonalPrint(root, k));
</script>
Producción:
4
Complejidad temporal : O(N), donde N es el número total de Nodes en el árbol binario.
Espacio Auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por Sakshi_Srivastava y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA