Sum of all the Boundary Nodes of a Binary Tree - GeeksforGeeks

28 Jun, 2021
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++
Java
Python3
C#
Javascript
// 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 childstruct Node {    int data;    struct Node* left;    struct Node* right;}; // Utility function to create a nodeNode* 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 nodesvoid 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 nodesvoid 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 treevoid 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 treeint 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 codeint 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 implementation of the approachclass GFG{    static int sum_of_boundary_nodes=0; // A binary tree node has data,// pointer to left childstatic class Node{    int data;    Node left;    Node right;}; // Utility function to create a nodestatic 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 nodesstatic 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 nodesstatic 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 treestatic 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 treestatic 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 codepublic 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 implementation of the approach  # A binary tree node has data,# pointer to left child# and a pointer to right childclass Node:         def __init__(self):                 self.left = None        self.right = None         sum_of_boundary_nodes = 0 # Utility function to create a nodedef newNode(data):     temp = Node()    temp.data = data;    return temp; # Function to sum up all the# left boundary nodes except# the leaf nodesdef 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 nodesdef 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 treedef 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 treedef 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 codeif __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# implementation of the approachusing System;class GFG{static int sum_of_boundary_nodes = 0; // A binary tree node has data,// pointer to left childpublic class Node{    public int data;    public Node left;    public Node right;}; // Utility function to create a nodestatic 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 nodesstatic 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 nodesstatic 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 treestatic 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 treestatic 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 codepublic 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
<script>     // JavaScript implementation of the approach         let sum_of_boundary_nodes=0;         // Binary Tree Node    class Node    {        constructor(data) {           this.left = null;           this.right = null;           this.data = data;        }    }     // Utility function to create a node    function newNode(data)    {        let temp = new Node(data);        return temp;    }     // Function to sum up all the left boundary nodes    // except the leaf nodes    function LeftBoundary(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    function RightBoundary(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    function Leaves(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    function sumOfBoundaryNodes(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;    }         let 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);      document.write(sumOfBoundaryNodes(root)); </script>
48
 
Time Complexity: O(N) where N is the number of nodes in the binary tree. 
andrew1234
princi singh
rutvik_56
mukesh07
Binary Tree
tree-traversal
Data Structures
Tree
Data Structures
Tree
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