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
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