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Print all full nodes in a Binary Tree

  • Difficulty Level : Easy
  • Last Updated : 22 Jun, 2021

Given a binary tree, print all nodes will are full nodes. Full Nodes are nodes which has both left and right children as non-empty.
Examples: 
 

Input :    10
          /  \
         8    2
        / \   /
       3   5 7
Output : 10 8

Input :   1
         / \
        2   3
           / \
          4   6     
Output : 1 3

 

This is a simple problem. We do any of the tra­ver­sals (Inorder, Pre­order, Pos­torder, level order traversal) and keep printing nodes that have mode left and right children as non-NULL.
 

C++




// A C++ program to find the all full nodes in
// a given binary tree
#include <bits/stdc++.h>
using namespace std;
 
struct Node
{
    int data;
    struct Node *left, *right;
};
 
Node *newNode(int data)
{
    Node *temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Traverses given tree in Inorder fashion and
// prints all nodes that have both children as
// non-empty.
void findFullNode(Node *root)
{
    if (root != NULL)
    {
        findFullNode(root->left);
        if (root->left != NULL && root->right != NULL)
            cout << root->data << " ";
        findFullNode(root->right);
    }
}
 
// Driver program to test above function
int main()
{
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(6);
    root->right->left->right = newNode(7);
    root->right->right->right = newNode(8);
    root->right->left->right->left = newNode(9);
    findFullNode(root);
    return 0;
}

Java




// Java program to find the all full nodes in
// a given binary tree
public class FullNodes {
 
    // Traverses given tree in Inorder fashion and
    // prints all nodes that have both children as
    // non-empty.
    public static void findFullNode(Node root)
    {
        if (root != null)
        {
            findFullNode(root.left);
            if (root.left != null && root.right != null)
                System.out.print(root.data+" ");
            findFullNode(root.right);
        }
    }
 
    public static void main(String args[]) {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.right.left = new Node(5);
        root.right.right = new Node(6);
        root.right.left.right = new Node(7);
        root.right.right.right = new Node(8);
        root.right.left.right.left = new Node(9);
        findFullNode(root);
    }
}
 
/* A binary tree node */
class Node
{
    int data;
    Node left, right;
    Node(int data)
    {
        left=right=null;
        this.data=data;
    }
};
//This code is contributed by Gaurav Tiwari

Python3




# Python3 program to find the all
# full nodes in a given binary tree
 
# Binary Tree Node
""" utility that allocates a newNode
with the given key """
class newNode:
 
    # Construct to create a newNode
    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None
 
# Traverses given tree in Inorder
# fashion and prints all nodes that
# have both children as non-empty.
def findFullNode(root) :
 
    if (root != None) :
     
        findFullNode(root.left)
        if (root.left != None and
            root.right != None) :
            print(root.data, end = " ")
        findFullNode(root.right)
 
# Driver Code
if __name__ == '__main__':
 
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.right.left = newNode(5)
    root.right.right = newNode(6)
    root.right.left.right = newNode(7)
    root.right.right.right = newNode(8)
    root.right.left.right.left = newNode(9)
    findFullNode(root)
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#




// C# program to find the all full nodes in
// a given binary tree
using System;
 
public class FullNodes
{
 
    // Traverses given tree in Inorder fashion and
    // prints all nodes that have both children as
    // non-empty.
    static void findFullNode(Node root)
    {
        if (root != null)
        {
            findFullNode(root.left);
            if (root.left != null && root.right != null)
                Console.Write(root.data + " ");
            findFullNode(root.right);
        }
    }
 
    public static void Main(String []args)
    {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.right.left = new Node(5);
        root.right.right = new Node(6);
        root.right.left.right = new Node(7);
        root.right.right.right = new Node(8);
        root.right.left.right.left = new Node(9);
        findFullNode(root);
    }
}
 
/* A binary tree node */
class Node
{
    public int data;
    public Node left, right;
    public Node(int data)
    {
        left = right = null;
        this.data = data;
    }
};
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript program to find the all full nodes in
// a given binary tree
 
/* A binary tree node */
class Node
{
    constructor(data)
    {
        this.left=this.right=null;
        this.data=data;
    }
}
 
// Traverses given tree in Inorder fashion and
    // prints all nodes that have both children as
    // non-empty.
function findFullNode(root)
{
    if (root != null)
        {
            findFullNode(root.left);
            if (root.left != null && root.right != null)
                document.write(root.data+" ");
            findFullNode(root.right);
        }
}
 
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.right.left = new Node(5);
root.right.right = new Node(6);
root.right.left.right = new Node(7);
root.right.right.right = new Node(8);
root.right.left.right.left = new Node(9);
findFullNode(root);
 
 
// This code is contributed by rag2127
 
</script>

Output:  

1 3

Time Complexity : O(n)
This article is contributed by Rakesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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