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Sum of all the Boundary Nodes of a Binary Tree
  • Difficulty Level : Hard
  • Last Updated : 22 Oct, 2020

Given a binary tree, the task is to print the sum of all the boundary nodes of the tree.
 

Examples: 

Input:
               1
             /   \
            2     3
           / \   / \
          4   5 6   7
Output: 28

Input:
                1
              /   \
             2     3
              \    /
               4  5
                  \
                   6
                  / \
                 7   8
Output: 36


Approach: We have already discussed the Boundary Traversal of a Binary tree. Here we will find the sum of the boundary nodes of the given binary tree in four steps: 

  • Sum up all the nodes of the left boundary,
  • Sum up all the leaf nodes of the left sub-tree,
  • Sum up all the leaf nodes of the right sub-tree and
  • Sum up all the nodes of the right boundary.

We will have to take care of one thing that nodes don’t add up again, i.e. the left most node is also the leaf node of the tree.



Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// A binary tree node has data,
// pointer to left child
// and a pointer to right child
struct Node {
    int data;
    struct Node* left;
    struct Node* right;
};
 
// Utility function to create a node
Node* newNode(int data)
{
    Node* temp = new Node;
 
    temp->left = NULL;
    temp->right = NULL;
    temp->data = data;
 
    return temp;
}
 
// Function to sum up all the left boundary nodes
// except the leaf nodes
void LeftBoundary(Node* root, int& sum_of_boundary_nodes)
{
    if (root) {
        if (root->left) {
            sum_of_boundary_nodes += root->data;
            LeftBoundary(root->left, sum_of_boundary_nodes);
        }
        else if (root->right) {
            sum_of_boundary_nodes += root->data;
            LeftBoundary(root->right, sum_of_boundary_nodes);
        }
    }
}
 
// Function to sum up all the right boundary nodes
// except the leaf nodes
void RightBoundary(Node* root, int& sum_of_boundary_nodes)
{
    if (root) {
        if (root->right) {
            RightBoundary(root->right, sum_of_boundary_nodes);
            sum_of_boundary_nodes += root->data;
        }
        else if (root->left) {
            RightBoundary(root->left, sum_of_boundary_nodes);
            sum_of_boundary_nodes += root->data;
        }
    }
}
 
// Function to sum up all the leaf nodes
// of a binary tree
void Leaves(Node* root, int& sum_of_boundary_nodes)
{
    if (root) {
        Leaves(root->left, sum_of_boundary_nodes);
 
        // Sum it up if it is a leaf node
        if (!(root->left) && !(root->right))
            sum_of_boundary_nodes += root->data;
 
        Leaves(root->right, sum_of_boundary_nodes);
    }
}
 
// Function to return the sum of all the
// boundary nodes of the given binary tree
int sumOfBoundaryNodes(struct Node* root)
{
    if (root) {
 
        // Root node is also a boundary node
        int sum_of_boundary_nodes = root->data;
 
        // Sum up all the left nodes
        // in TOP DOWN manner
        LeftBoundary(root->left, sum_of_boundary_nodes);
 
        // Sum up all the
        // leaf nodes
        Leaves(root->left, sum_of_boundary_nodes);
        Leaves(root->right, sum_of_boundary_nodes);
 
        // Sum up all the right nodes
        // in BOTTOM UP manner
        RightBoundary(root->right, sum_of_boundary_nodes);
 
        // Return the sum of
        // all the boundary nodes
        return sum_of_boundary_nodes;
    }
 
    return 0;
}
 
// Driver code
int main()
{
    Node* root = newNode(10);
    root->left = newNode(2);
    root->right = newNode(5);
    root->left->left = newNode(8);
    root->left->right = newNode(14);
    root->right->left = newNode(11);
    root->right->right = newNode(3);
    root->left->right->left = newNode(12);
    root->right->left->right = newNode(1);
    root->right->left->left = newNode(7);
 
    cout << sumOfBoundaryNodes(root);
 
    return 0;
}


Java




// Java implementation of the approach
class GFG
{
    static int sum_of_boundary_nodes=0;
 
// A binary tree node has data,
// pointer to left child
static class Node
{
    int data;
    Node left;
    Node right;
};
 
// Utility function to create a node
static Node newNode(int data)
{
    Node temp = new Node();
 
    temp.left = null;
    temp.right = null;
    temp.data = data;
 
    return temp;
}
 
// Function to sum up all the left boundary nodes
// except the leaf nodes
static void LeftBoundary(Node root)
{
    if (root != null)
    {
        if (root.left != null)
        {
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.left);
        }
        else if (root.right != null)
        {
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.right);
        }
    }
}
 
// Function to sum up all the right boundary nodes
// except the leaf nodes
static void RightBoundary(Node root)
{
    if (root != null)
    {
        if (root.right != null)
        {
            RightBoundary(root.right);
            sum_of_boundary_nodes += root.data;
        }
        else if (root.left != null)
        {
            RightBoundary(root.left);
            sum_of_boundary_nodes += root.data;
        }
    }
}
 
// Function to sum up all the leaf nodes
// of a binary tree
static void Leaves(Node root)
{
    if (root != null)
    {
        Leaves(root.left);
 
        // Sum it up if it is a leaf node
        if ((root.left == null) && (root.right == null))
            sum_of_boundary_nodes += root.data;
 
        Leaves(root.right);
    }
}
 
// Function to return the sum of all the
// boundary nodes of the given binary tree
static int sumOfBoundaryNodes( Node root)
{
    if (root != null)
    {
 
        // Root node is also a boundary node
        sum_of_boundary_nodes = root.data;
 
        // Sum up all the left nodes
        // in TOP DOWN manner
        LeftBoundary(root.left);
 
        // Sum up all the
        // leaf nodes
        Leaves(root.left);
        Leaves(root.right);
 
        // Sum up all the right nodes
        // in BOTTOM UP manner
        RightBoundary(root.right);
 
        // Return the sum of
        // all the boundary nodes
        return sum_of_boundary_nodes;
    }
 
    return 0;
}
 
// Driver code
public static void main(String args[])
{
    Node root = newNode(10);
    root.left = newNode(2);
    root.right = newNode(5);
    root.left.left = newNode(8);
    root.left.right = newNode(14);
    root.right.left = newNode(11);
    root.right.right = newNode(3);
    root.left.right.left = newNode(12);
    root.right.left.right = newNode(1);
    root.right.left.left = newNode(7);
 
    System.out.println(sumOfBoundaryNodes(root));
}
}
 
// This code is contributed by andrew1234


Python3




# Python3 implementation of the approach
  
# A binary tree node has data,
# pointer to left child
# and a pointer to right child
class Node:
     
    def __init__(self):
         
        self.left = None
        self.right = None
         
sum_of_boundary_nodes = 0
 
# Utility function to create a node
def newNode(data):
 
    temp = Node()
    temp.data = data;
    return temp;
 
# Function to sum up all the
# left boundary nodes except
# the leaf nodes
def LeftBoundary(root):
     
    global sum_of_boundary_nodes
     
    if (root != None):
        if (root.left != None):
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.left);
         
        elif (root.right != None):
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.right);
 
# Function to sum up all the right
# boundary nodes except the leaf nodes
def RightBoundary(root):
     
    global sum_of_boundary_nodes
     
    if (root != None):
        if (root.right != None):
            RightBoundary(root.right);
            sum_of_boundary_nodes += root.data;
         
        elif (root.left != None):
            RightBoundary(root.left);
            sum_of_boundary_nodes += root.data;
         
# Function to sum up all the leaf nodes
# of a binary tree
def Leaves(root):
     
    global sum_of_boundary_nodes
     
    if (root != None):
        Leaves(root.left);
  
        # Sum it up if it is a leaf node
        if ((root.left == None) and
           (root.right == None)):
            sum_of_boundary_nodes += root.data;
  
        Leaves(root.right);
     
# Function to return the sum of all the
# boundary nodes of the given binary tree
def sumOfBoundaryNodes(root):
     
    global sum_of_boundary_nodes
     
    if (root != None):
         
        # Root node is also a boundary node
        sum_of_boundary_nodes = root.data;
  
        # Sum up all the left nodes
        # in TOP DOWN manner
        LeftBoundary(root.left);
  
        # Sum up all the
        # leaf nodes
        Leaves(root.left);
        Leaves(root.right);
  
        # Sum up all the right nodes
        # in BOTTOM UP manner
        RightBoundary(root.right);
  
        # Return the sum of
        # all the boundary nodes
        return sum_of_boundary_nodes;
     
    return 0;
 
# Driver code
if __name__=="__main__":
     
    root = newNode(10);
    root.left = newNode(2);
    root.right = newNode(5);
    root.left.left = newNode(8);
    root.left.right = newNode(14);
    root.right.left = newNode(11);
    root.right.right = newNode(3);
    root.left.right.left = newNode(12);
    root.right.left.right = newNode(1);
    root.right.left.left = newNode(7);
  
    print(sumOfBoundaryNodes(root));
 
# This code is contributed by rutvik_56


C#




// C# implementation of the approach
using System;
class GFG
{
static int sum_of_boundary_nodes = 0;
 
// A binary tree node has data,
// pointer to left child
public class Node
{
    public int data;
    public Node left;
    public Node right;
};
 
// Utility function to create a node
static Node newNode(int data)
{
    Node temp = new Node();
 
    temp.left = null;
    temp.right = null;
    temp.data = data;
 
    return temp;
}
 
// Function to sum up all the left boundary
// nodes except the leaf nodes
static void LeftBoundary(Node root)
{
    if (root != null)
    {
        if (root.left != null)
        {
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.left);
        }
         
        else if (root.right != null)
        {
            sum_of_boundary_nodes += root.data;
            LeftBoundary(root.right);
        }
    }
}
 
// Function to sum up all the right boundary
// nodes except the leaf nodes
static void RightBoundary(Node root)
{
    if (root != null)
    {
        if (root.right != null)
        {
            RightBoundary(root.right);
            sum_of_boundary_nodes += root.data;
        }
        else if (root.left != null)
        {
            RightBoundary(root.left);
            sum_of_boundary_nodes += root.data;
        }
    }
}
 
// Function to sum up all the leaf nodes
// of a binary tree
static void Leaves(Node root)
{
    if (root != null)
    {
        Leaves(root.left);
 
        // Sum it up if it is a leaf node
        if ((root.left == null) &&
            (root.right == null))
            sum_of_boundary_nodes += root.data;
 
        Leaves(root.right);
    }
}
 
// Function to return the sum of all the
// boundary nodes of the given binary tree
static int sumOfBoundaryNodes(Node root)
{
    if (root != null)
    {
 
        // Root node is also a boundary node
        sum_of_boundary_nodes = root.data;
 
        // Sum up all the left nodes
        // in TOP DOWN manner
        LeftBoundary(root.left);
 
        // Sum up all the
        // leaf nodes
        Leaves(root.left);
        Leaves(root.right);
 
        // Sum up all the right nodes
        // in BOTTOM UP manner
        RightBoundary(root.right);
 
        // Return the sum of
        // all the boundary nodes
        return sum_of_boundary_nodes;
    }
    return 0;
}
 
// Driver code
public static void Main(String []args)
{
    Node root = newNode(10);
    root.left = newNode(2);
    root.right = newNode(5);
    root.left.left = newNode(8);
    root.left.right = newNode(14);
    root.right.left = newNode(11);
    root.right.right = newNode(3);
    root.left.right.left = newNode(12);
    root.right.left.right = newNode(1);
    root.right.left.left = newNode(7);
 
    Console.WriteLine(sumOfBoundaryNodes(root));
}
}
 
// This code is contributed by Princi Singh


Output: 

48


 

Time Complexity: O(N) where N is the number of nodes in the binary tree.
 

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