Find n-th node of inorder traversal
Given the binary tree and you have to find out the n-th node of inorder traversal.
Examples:
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Input : n = 4
10
/ \
20 30
/ \
40 50
Output : 10
Inorder Traversal is : 40 20 50 10 30
Input : n = 3
7
/ \
2 3
/ \
8 5
Output : 8
Inorder: 2 7 8 3 5
3th node is 8
We do simple Inorder Traversal. While doing the traversal, we keep track of count of nodes visited so far. When count becomes n, we print the node.
Below is the implementation of the above approach.
C
// C program for nth nodes of inorder traversals#include <stdio.h>#include <stdlib.h>/* A binary tree node has data, pointer to left childand a pointer to right child */struct Node { int data; struct Node* left; struct Node* right;};/* Helper function that allocates a new node with thegiven 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 = Falsecount = [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 prelement if (count[0] == n): print(node.data, end = " ") """ now recur on right child """ NthInorder(node.right, n) # Driver Codeif __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 Codepublic 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> |
Output:
10
Time Complexity: O(n)
