Hello Everyone,
Given the binary tree and you have to find out the n-th node of inorder traversal.
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 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;
}
Output:
10
Time Complexity: O(n)