Print all full nodes in a Binary Tree
Last Updated :
13 Sep, 2023
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 traÂverÂsals (Inorder, PreÂorder, PosÂtorder, level order traversal) and keep printing nodes that have mode left and right children as non-NULL.
Implementation:
C++
#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;
}
void findFullNode(Node *root)
{
if (root != NULL)
{
findFullNode(root->left);
if (root->left != NULL && root->right != NULL)
cout << root->data << " " ;
findFullNode(root->right);
}
}
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
public class FullNodes {
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);
}
}
class Node
{
int data;
Node left, right;
Node( int data)
{
left=right= null ;
this .data=data;
}
};
|
Python3
class newNode:
def __init__( self , key):
self .data = key
self .left = None
self .right = None
def findFullNode(root) :
if (root ! = None ) :
findFullNode(root.left)
if (root.left ! = None and
root.right ! = None ) :
print (root.data, end = " " )
findFullNode(root.right)
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)
|
C#
using System;
public class FullNodes
{
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);
}
}
class Node
{
public int data;
public Node left, right;
public Node( int data)
{
left = right = null ;
this .data = data;
}
};
|
Javascript
<script>
class Node
{
constructor(data)
{
this .left= this .right= null ;
this .data=data;
}
}
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);
</script>
|
Time Complexity : O(n)
Space complexity: O(n) for Recursive Stack Space in case of Skewed Tree
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