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Print all full nodes in a Binary Tree

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  • Difficulty Level : Easy
  • Last Updated : 08 Jul, 2022
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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.

Implementation:

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)
Space complexity: O(n) for Recursive Stack Space in case of Skewed Tree

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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