Given a binary tree, you need to check whether sum of all covered elements is equal to sum of all uncovered elements or not.
In a binary tree, a node is called Uncovered if it appears either on left boundary or right boundary. Rest of the nodes are called covered.
For example, consider below binary tree
In above binary tree, Covered node: 6, 5, 7 Uncovered node: 9, 4, 3, 17, 22, 20 The output for this tree should be false as sum of covered and uncovered node is not same
We strongly recommend you to minimize your browser and try this yourself first.
For calculating sum of Uncovered nodes we will follow below steps:
1) Start from root, go to left and keep going until left child is available, if not go to right child and again follow same procedure until you reach a leaf node.
2) After step 1 sum of left boundary will be stored, now for right part again do the same procedure but now keep going to right until right child is available, if not then go to left child and follow same procedure until you reach a leaf node.
After above 2 steps sum of all Uncovered node will be stored, we can subtract it from total sum and get sum of covered elements and check for equines of binary tree.
// C++ program to find sum of Covered and Uncovered node of // binary tree #include <bits/stdc++.h> using namespace std;
/* A binary tree node has key, pointer to left child and a pointer to right child */
struct Node
{ int key;
struct Node* left, *right;
}; /* To create a newNode of tree and return pointer */ struct Node* newNode( int key)
{ Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return (temp);
} /* Utility function to calculate sum of all node of tree */ int sum(Node* t)
{ if (t == NULL)
return 0;
return t->key + sum(t->left) + sum(t->right);
} /* Recursive function to calculate sum of left boundary elements */
int uncoveredSumLeft(Node* t)
{ /* If leaf node, then just return its key value */
if (t->left == NULL && t->right == NULL)
return t->key;
/* If left is available then go left otherwise go right */
if (t->left != NULL)
return t->key + uncoveredSumLeft(t->left);
else
return t->key + uncoveredSumLeft(t->right);
} /* Recursive function to calculate sum of right boundary elements */
int uncoveredSumRight(Node* t)
{ /* If leaf node, then just return its key value */
if (t->left == NULL && t->right == NULL)
return t->key;
/* If right is available then go right otherwise go left */
if (t->right != NULL)
return t->key + uncoveredSumRight(t->right);
else
return t->key + uncoveredSumRight(t->left);
} // Returns sum of uncovered elements int uncoverSum(Node* t)
{ /* Initializing with 0 in case we don't have
left or right boundary */
int lb = 0, rb = 0;
if (t->left != NULL)
lb = uncoveredSumLeft(t->left);
if (t->right != NULL)
rb = uncoveredSumRight(t->right);
/* returning sum of root node, left boundary
and right boundary*/
return t->key + lb + rb;
} // Returns true if sum of covered and uncovered elements // is same. bool isSumSame(Node *root)
{ // Sum of uncovered elements
int sumUC = uncoverSum(root);
// Sum of all elements
int sumT = sum(root);
// Check if sum of covered and uncovered is same
return (sumUC == (sumT - sumUC));
} /* Helper function to print inorder traversal of binary tree */
void inorder(Node* root)
{ if (root)
{
inorder(root->left);
printf ( "%d " , root->key);
inorder(root->right);
}
} // Driver program to test above functions int main()
{ // Making above given diagram's binary tree
Node* root = newNode(8);
root->left = newNode(3);
root->left->left = newNode(1);
root->left->right = newNode(6);
root->left->right->left = newNode(4);
root->left->right->right = newNode(7);
root->right = newNode(10);
root->right->right = newNode(14);
root->right->right->left = newNode(13);
if (isSumSame(root))
printf ( "Sum of covered and uncovered is same\n" );
else
printf ( "Sum of covered and uncovered is not same\n" );
} |
// Java program to find sum of covered and uncovered nodes // of a binary tree /* A binary tree node has key, pointer to left child and a pointer to right child */
class Node
{ int key;
Node left, right;
public Node( int key)
{
this .key = key;
left = right = null ;
}
} class BinaryTree
{ Node root;
/* Utility function to calculate sum of all node of tree */
int sum(Node t)
{
if (t == null )
return 0 ;
return t.key + sum(t.left) + sum(t.right);
}
/* Recursive function to calculate sum of left boundary
elements */
int uncoveredSumLeft(Node t)
{
/* If left node, then just return its key value */
if (t.left == null && t.right == null )
return t.key;
/* If left is available then go left otherwise go right */
if (t.left != null )
return t.key + uncoveredSumLeft(t.left);
else
return t.key + uncoveredSumLeft(t.right);
}
/* Recursive function to calculate sum of right boundary
elements */
int uncoveredSumRight(Node t)
{
/* If left node, then just return its key value */
if (t.left == null && t.right == null )
return t.key;
/* If right is available then go right otherwise go left */
if (t.right != null )
return t.key + uncoveredSumRight(t.right);
else
return t.key + uncoveredSumRight(t.left);
}
// Returns sum of uncovered elements
int uncoverSum(Node t)
{
/* Initializing with 0 in case we don't have
left or right boundary */
int lb = 0 , rb = 0 ;
if (t.left != null )
lb = uncoveredSumLeft(t.left);
if (t.right != null )
rb = uncoveredSumRight(t.right);
/* returning sum of root node, left boundary
and right boundary*/
return t.key + lb + rb;
}
// Returns true if sum of covered and uncovered elements
// is same.
boolean isSumSame(Node root)
{
// Sum of uncovered elements
int sumUC = uncoverSum(root);
// Sum of all elements
int sumT = sum(root);
// Check if sum of covered and uncovered is same
return (sumUC == (sumT - sumUC));
}
/* Helper function to print inorder traversal of
binary tree */
void inorder(Node root)
{
if (root != null )
{
inorder(root.left);
System.out.print(root.key + " " );
inorder(root.right);
}
}
// Driver program to test above functions
public static void main(String[] args)
{
BinaryTree tree = new BinaryTree();
// Making above given diagram's binary tree
tree.root = new Node( 8 );
tree.root.left = new Node( 3 );
tree.root.left.left = new Node( 1 );
tree.root.left.right = new Node( 6 );
tree.root.left.right.left = new Node( 4 );
tree.root.left.right.right = new Node( 7 );
tree.root.right = new Node( 10 );
tree.root.right.right = new Node( 14 );
tree.root.right.right.left = new Node( 13 );
if (tree.isSumSame(tree.root))
System.out.println( "Sum of covered and uncovered is same" );
else
System.out.println( "Sum of covered and uncovered is not same" );
}
} // This code has been contributed by Mayank Jaiswal(mayank_24) |
# Python3 program to find sum of Covered and # Uncovered node of binary tree # To create a newNode of tree and return pointer class newNode:
def __init__( self , key):
self .key = key
self .left = self .right = None
# Utility function to calculate sum # of all node of tree def Sum (t):
if (t = = None ):
return 0
return t.key + Sum (t.left) + Sum (t.right)
# Recursive function to calculate sum # of left boundary elements def uncoveredSumLeft(t):
# If leaf node, then just return
# its key value
if (t.left = = None and t.right = = None ):
return t.key
# If left is available then go
# left otherwise go right
if (t.left ! = None ):
return t.key + uncoveredSumLeft(t.left)
else :
return t.key + uncoveredSumLeft(t.right)
# Recursive function to calculate sum of # right boundary elements def uncoveredSumRight(t):
# If leaf node, then just return
# its key value
if (t.left = = None and t.right = = None ):
return t.key
# If right is available then go right
# otherwise go left
if (t.right ! = None ):
return t.key + uncoveredSumRight(t.right)
else :
return t.key + uncoveredSumRight(t.left)
# Returns sum of uncovered elements def uncoverSum(t):
# Initializing with 0 in case we
# don't have left or right boundary
lb = 0
rb = 0
if (t.left ! = None ):
lb = uncoveredSumLeft(t.left)
if (t.right ! = None ):
rb = uncoveredSumRight(t.right)
# returning sum of root node,
# left boundary and right boundary
return t.key + lb + rb
# Returns true if sum of covered and # uncovered elements is same. def isSumSame(root):
# Sum of uncovered elements
sumUC = uncoverSum(root)
# Sum of all elements
sumT = Sum (root)
# Check if sum of covered and
# uncovered is same
return (sumUC = = (sumT - sumUC))
# Helper function to print Inorder # traversal of binary tree def inorder(root):
if (root):
inorder(root.left)
print (root.key, end = " " )
inorder(root.right)
# Driver Code if __name__ = = '__main__' :
# Making above given diagram's
# binary tree
root = newNode( 8 )
root.left = newNode( 3 )
root.left.left = newNode( 1 )
root.left.right = newNode( 6 )
root.left.right.left = newNode( 4 )
root.left.right.right = newNode( 7 )
root.right = newNode( 10 )
root.right.right = newNode( 14 )
root.right.right.left = newNode( 13 )
if (isSumSame(root)):
print ( "Sum of covered and uncovered is same" )
else :
print ( "Sum of covered and uncovered is not same" )
# This code is contributed by PranchalK |
// C# program to find sum of covered // and uncovered nodes of a binary tree using System;
/* A binary tree node has key, pointer to left child and a pointer to right child */ public class Node
{ public int key;
public Node left, right;
public Node( int key)
{
this .key = key;
left = right = null ;
}
} class GFG
{ public Node root;
/* Utility function to calculate sum of all node of tree */ public virtual int sum(Node t)
{ if (t == null )
{
return 0;
}
return t.key + sum(t.left) +
sum(t.right);
} /* Recursive function to calculate sum of left boundary elements */ public virtual int uncoveredSumLeft(Node t)
{ /* If left node, then just return
its key value */
if (t.left == null && t.right == null )
{
return t.key;
}
/* If left is available then go
left otherwise go right */
if (t.left != null )
{
return t.key + uncoveredSumLeft(t.left);
}
else
{
return t.key + uncoveredSumLeft(t.right);
}
} /* Recursive function to calculate sum of right boundary elements */
public virtual int uncoveredSumRight(Node t)
{ /* If left node, then just return
its key value */
if (t.left == null && t.right == null )
{
return t.key;
}
/* If right is available then go
right otherwise go left */
if (t.right != null )
{
return t.key + uncoveredSumRight(t.right);
}
else
{
return t.key + uncoveredSumRight(t.left);
}
} // Returns sum of uncovered elements public virtual int uncoverSum(Node t)
{ /* Initializing with 0 in case we
don't have left or right boundary */
int lb = 0, rb = 0;
if (t.left != null )
{
lb = uncoveredSumLeft(t.left);
}
if (t.right != null )
{
rb = uncoveredSumRight(t.right);
}
/* returning sum of root node,
left boundary and right boundary*/
return t.key + lb + rb;
} // Returns true if sum of covered // and uncovered elements is same. public virtual bool isSumSame(Node root)
{ // Sum of uncovered elements
int sumUC = uncoverSum(root);
// Sum of all elements
int sumT = sum(root);
// Check if sum of covered and
// uncovered is same
return (sumUC == (sumT - sumUC));
} /* Helper function to print inorder traversal of binary tree */ public virtual void inorder(Node root)
{ if (root != null )
{
inorder(root.left);
Console.Write(root.key + " " );
inorder(root.right);
}
} // Driver Code public static void Main( string [] args)
{ GFG tree = new GFG();
// Making above given diagram's binary tree
tree.root = new Node(8);
tree.root.left = new Node(3);
tree.root.left.left = new Node(1);
tree.root.left.right = new Node(6);
tree.root.left.right.left = new Node(4);
tree.root.left.right.right = new Node(7);
tree.root.right = new Node(10);
tree.root.right.right = new Node(14);
tree.root.right.right.left = new Node(13);
if (tree.isSumSame(tree.root))
{
Console.WriteLine( "Sum of covered and " +
"uncovered is same" );
}
else
{
Console.WriteLine( "Sum of covered and " +
"uncovered is not same" );
}
} } // This code is contributed by Shrikant13 |
<script> // Javascript program to find sum of covered // and uncovered nodes of a binary tree /* A binary tree node has key, pointer to left child and a pointer to right child */ class Node { constructor(key)
{
this .key = key;
this .left= null ;
this .right = null ;
}
} var root = null ;
/* Utility function to calculate sum of all node of tree */ function sum(t)
{ if (t == null )
{
return 0;
}
return t.key + sum(t.left) +
sum(t.right);
} /* Recursive function to calculate sum of left boundary elements */ function uncoveredSumLeft(t)
{ /* If left node, then just return
its key value */
if (t.left == null && t.right == null )
{
return t.key;
}
/* If left is available then go
left otherwise go right */
if (t.left != null )
{
return t.key + uncoveredSumLeft(t.left);
}
else
{
return t.key + uncoveredSumLeft(t.right);
}
} /* Recursive function to calculate sum of right boundary elements */
function uncoveredSumRight(t)
{ /* If left node, then just return
its key value */
if (t.left == null && t.right == null )
{
return t.key;
}
/* If right is available then go
right otherwise go left */
if (t.right != null )
{
return t.key + uncoveredSumRight(t.right);
}
else
{
return t.key + uncoveredSumRight(t.left);
}
} // Returns sum of uncovered elements function uncoverSum(t)
{ /* Initializing with 0 in case we
don't have left or right boundary */
var lb = 0, rb = 0;
if (t.left != null )
{
lb = uncoveredSumLeft(t.left);
}
if (t.right != null )
{
rb = uncoveredSumRight(t.right);
}
/* returning sum of root node,
left boundary and right boundary*/
return t.key + lb + rb;
} // Returns true if sum of covered // and uncovered elements is same. function isSumSame(root)
{ // Sum of uncovered elements
var sumUC = uncoverSum(root);
// Sum of all elements
var sumT = sum(root);
// Check if sum of covered and
// uncovered is same
return (sumUC == (sumT - sumUC));
} /* Helper function to print inorder traversal of binary tree */ function inorder(root)
{ if (root != null )
{
inorder(root.left);
document.write(root.key + " " );
inorder(root.right);
}
} // Driver Code // Making above given diagram's binary tree var root = new Node(8);
root.left = new Node(3);
root.left.left = new Node(1);
root.left.right = new Node(6);
root.left.right.left = new Node(4);
root.left.right.right = new Node(7);
root.right = new Node(10);
root.right.right = new Node(14);
root.right.right.left = new Node(13);
if (isSumSame(root))
{ document.write( "Sum of covered and " +
"uncovered is same" );
} else { document.write( "Sum of covered and " +
"uncovered is not same" );
} // This code is contributed by itsok </script> |
Output :
Sum of covered and uncovered is not same
Time Complexity: O(N), node will run for every new node.
Space Complexity: O(N), space is needed for every node.