Print the nodes at odd levels of a tree in C++ Programming.

Given the binary tree, the program must print the nodes at odd levels of a tree and the levels of a binary tree start from 1 to n.

As nothing is mentioned one of the two approaches can be implemented i.e. recursion or iteration.

Since we are using a recursive approach, the program will make a recursive call to a function that will be fetching the nodes at odd levels and returning them.

In the above binary tree −

Nodes at level 1: 10
Nodes at level 2: 3 and 211
Nodes at level 3: 140, 162, 100 and 146

So, the nodes at level 1 and level 3 will be printed that means the output will be 10, 140, 162, 100 and 146.


Step 1 -> create a structure of a node as
   struct Node
      struct node *left, *right
      int data
Step 2 -> function to create a node
   node* newnode(int data)
   node->data = data
   node->left = node->right = NULL;
   return (node)
step 3 -> create function for finding the odd nodes
   void odd(Node *root, bool ifodd = true)
   IF root = NULL
   if (ifodd)
      print root->data
   odd(root->left, !ifodd)
   odd(root->right, !ifodd)
step 4 -> In main()
   Create tree using Node* root = newnode(45)
   root->left = newnode(23)
   Call odd(root)


 Live Demo

#include <bits/stdc++.h>
using namespace std;
struct Node{
   int data;
   Node* left, *right;
void odd(Node *root, bool ifodd = true){
   if (root == NULL)
   if (ifodd)
      cout << root->data << " " ;
   odd(root->left, !ifodd);
   odd(root->right, !ifodd);
// function to create a new node
Node* newnode(int data){
   Node* node = new Node;
   node->data = data;
   node->left = node->right = NULL;
   return (node);
int main(){
   Node* root = newnode(45);
   root->left = newnode(23);
   root->right = newnode(13);
   root->left->left = newnode(24);
   root->left->right = newnode(85);
   cout<<"\nodd nodes are ";
   return 0;


if we run the above program then it will generate the following output

odd nodes are 45 24 85

Updated on: 04-Sep-2019


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