# Find n-th node of inorder traversal

• Difficulty Level : Basic
• Last Updated : 20 Dec, 2022

Given the binary tree and you have to find out the n-th node of inorder traversal.

Examples:

```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 the count of nodes visited so far. When the 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 ` `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;` `    ``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)` `            ``cout << node->data<< endl;`   `        ``/* 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;` `}`   `// This code is contributed by Abhijeet Kumar(abhijeet19403)`

## C

 `// C program for  nth nodes of  inorder traversals` `#include ` `#include `   `/* 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.io.*;` `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

 ``

Output

```10
```

Time Complexity: O(N), where N is the number of nodes in the given binary tree. We need to traverse the given binary tree to find the nth node of inorder traversal.
Auxiliary Space: O(1), we only need to maintain a static variable to store the count of the node while traversing the binary tree.

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