Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Check sum of Covered and Uncovered nodes of Binary Tree

  • Difficulty Level : Medium
  • Last Updated : 25 Jun, 2021

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

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++




// 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




// 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




# 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 prinorder
# 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#




// 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

Javascript




<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
 

 

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!