Dado el árbol binario, debe encontrar el n-ésimo Node del recorrido en orden .
Ejemplos:
C
// C program for nth nodes of inorder traversals
#include <stdio.h>
#include <stdlib.h>
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
int data;
struct Node* left;
struct Node* right;
};
/* 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 =
(struct Node*)malloc(sizeof(struct Node));
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
/* Given a binary tree, print its nth nodes of inorder*/
void NthInorder(struct Node* node, int n)
{
static int count = 0;
if (node == NULL)
return;
if (count <= n) {
/* first recur on left child */
NthInorder(node->left, n);
count++;
// when count = n then print element
if (count == n)
printf("%d ", node->data);
/* now recur on right child */
NthInorder(node->right, n);
}
}
/* Driver program to test above functions*/
int main()
{
struct Node* root = newNode(10);
root->left = newNode(20);
root->right = newNode(30);
root->left->left = newNode(40);
root->left->right = newNode(50);
int n = 4;
NthInorder(root, n);
return 0;
}
Java
// Java program for nth nodes of inorder traversals
import java.util. *;
class Solution
{
static int count =0;
/* A binary tree node has data, pointer to left child
and a pointer to right child */
static class Node {
int data;
Node left;
Node right;
}
/* 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);
}
/* Given a binary tree, print its nth nodes of inorder*/
static void NthInorder( Node node, int n)
{
if (node == null)
return;
if (count <= n) {
/* first recur on left child */
NthInorder(node.left, n);
count++;
// when count = n then print element
if (count == n)
System.out.printf("%d ", node.data);
/* now recur on right child */
NthInorder(node.right, n);
}
}
/* Driver program to test above functions*/
public static void main(String args[])
{
Node root = newNode(10);
root.left = newNode(20);
root.right = newNode(30);
root.left.left = newNode(40);
root.left.right = newNode(50);
int n = 4;
NthInorder(root, n);
}
}
// This code is contributed
// by Arnab Kundu
Python3
"""Python3 program for nth node of inorder traversal""" # A Binary Tree Node # Utility function to create a # new tree node class newNode: # Constructor to create a newNode def __init__(self, data): self.data = data self.left = None self.right = None self.visited = False count = [0] """ Given a binary tree, print the nth node of inorder traversal""" def NthInorder(node, n): if (node == None): return if (count[0] <= n): """ first recur on left child """ NthInorder(node.left, n) count[0] += 1 # when count = n then print element if (count[0] == n): print(node.data, end = " ") """ now recur on right child """ NthInorder(node.right, n) # Driver Code if __name__ == '__main__': root = newNode(10) root.left = newNode(20) root.right = newNode(30) root.left.left = newNode(40) root.left.right = newNode(50) n = 4 NthInorder(root, n) # This code is contributed # by SHUBHAMSINGH10
C#
// C# program for nth nodes of
// inorder traversals
using System;
class GFG
{
public static int count = 0;
/* 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;
public Node right;
}
/* Helper function that allocates a
new node with the given data and null
left and right pointers. */
public static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
/* Given a binary tree, print its
nth nodes of inorder*/
public static void NthInorder(Node node, int n)
{
if (node == null)
{
return;
}
if (count <= n)
{
/* first recur on left child */
NthInorder(node.left, n);
count++;
// when count = n then print element
if (count == n)
{
Console.Write("{0:D} ", node.data);
}
/* now recur on right child */
NthInorder(node.right, n);
}
}
// Driver Code
public static void Main(string[] args)
{
Node root = newNode(10);
root.left = newNode(20);
root.right = newNode(30);
root.left.left = newNode(40);
root.left.right = newNode(50);
int n = 4;
NthInorder(root, n);
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// JavaScript program for nth nodes of inorder traversals
/* A binary tree node has data, pointer to left child
and a pointer to right child */
class Node
{
constructor(data)
{
this.data=data;
this.left=this.right=null;
}
}
let count =0;
/* Given a binary tree, print its nth nodes of inorder*/
function NthInorder(node,n)
{
if (node == null)
return;
if (count <= n) {
/* first recur on left child */
NthInorder(node.left, n);
count++;
// when count = n then print element
if (count == n)
document.write(node.data+" ");
/* now recur on right child */
NthInorder(node.right, n);
}
}
/* Driver program to test above functions*/
let root = new Node(10);
root.left = new Node(20);
root.right = new Node(30);
root.left.left = new Node(40);
root.left.right = new Node(50);
let n = 4;
NthInorder(root, n);
// This code is contributed by avanitrachhadiya2155
</script>
Publicación traducida automáticamente
Artículo escrito por shrikanth13 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA