Sum of all the Boundary Nodes of a Binary Tree

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:

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// 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;
}

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Output:

48

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



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