Print All Internal Nodes of a Binary Tree in C++

In this problem, we are given a binary tree and we have to print all internal nodes of the binary tree.

The binary tree is a tree in which a node can have a maximum of 2 child nodes. Node or vertex can have no nodes, one child or two child nodes.

Example −

Internal Node is a node that can have at least one child i.e. non-leaf node is an internal node.

Let’s take an example to understand the problem −

Output − 7 4 9

To solve this problem, we will traverse the binary tree using BFS(breadth-first search) traversal.

While traversal we will push nodes to a queue. When we pop elements from the queue, we will print all nodes of the tree that do not have any child nodes.

Example

Our logic is implemented by the below code −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
struct Node {
   int data;
   Node *left, *right;
   Node(int data){
      left = right = NULL;
      this->data = data;
   }
};
void printNonLeafNodes(Node* root) {
   queue<Node*> treeNodes;
   treeNodes.push(root);
   while (!treeNodes.empty()) {
      Node* curr = treeNodes.front();
      treeNodes.pop();
      bool isInternal = 0;
      if (curr->left) {
         isInternal = 1;
         treeNodes.push(curr->left);
      }
      if (curr->right) {
         isInternal = 1;
         treeNodes.push(curr->right);
      }
      if (isInternal)
         cout<<curr->data<<"\t";
   }
}
int main() {
   Node* root = new Node(43);
   root->left = new Node(12);
   root->right = new Node(78);
   root->left->left = new Node(4);
   root->right->left = new Node(9);
   root->right->right = new Node(1);
   root->right->right->right = new Node(50);
   root->right->right->left = new Node(25);
   cout<<"All internal Nodes of the binary tree are :\n";
   printNonLeafNodes(root);
   return 0;
}

Output

All internal Nodes of the binary tree are −
43 12 78 1