This is a C Program to find the sum of all the nodes present in a Binary Tree using recursion.

Problem Description

We have to write a C program which will find the sum of all the nodes in a Binary Tree.

Expected Input and Output

Case 1. Balanced Tree:When the weight is equal on both the sides of root.

25

/ \

27 19

/ \ / \

17 91 13 55

Output: 247

Case 2. Right Skewed Tree:When the nodes at every level just have a right child.

1

\

2

\

3

\

4

\

5

Output: 15

Subscribe Now: Data Structures in C Newsletter | Important Subjects Newsletters

Case 3. Tree having just one node

15

Output: 15

Participate in Data Structure I Certification Contest of the Month Now!

Problem Solution

There are two ways we can find the sum of all the nodes in a given Binary Tree.

Approach One

We can store the elements of tree in an array by using any of the traversal techniques and then find the sum of the array elements.

Approach Two

We keep on adding the values of nodes as we traverse them. We do this till we traverse the whole tree.

Here in this program we will be using second approach because if we use first approach we will have to create an array which will take up some space. We can avoid this by using second approach.

Program/Source Code

Here is source code of the C Program for finding the sum of all the nodes in a binary tree. The program is successfully compiled and tested using Codeblocks gnu/GCC compiler on windows 10. The program output is also shown below.

/* C Program for finding the sum of all the nodes in a Tree */

#include

#include

struct node

{

int info;

struct node *left, *right;

};

struct node *createnode(int key)

{

struct node *newnode = (struct node*)malloc(sizeof(struct node));

newnode->info = key;

newnode->left = NULL;

newnode->right = NULL;

return(newnode);

}

int sumofnodes(struct node *root)

{

int rightsubtree, leftsubtree, sum = 0;

if(root != NULL)

{

leftsubtree = sumofnodes(root->left);

rightsubtree = sumofnodes(root->right);

sum = (root->info) + leftsubtree + rightsubtree;

return sum;

}

}

/*

* Main Function

*/

int main()

{

/* Creating first Tree. */

struct node *newnode = createnode(25);

newnode->left = createnode(27);

newnode->right = createnode(19);

newnode->left->left = createnode(17);

newnode->left->right = createnode(91);

newnode->right->left = createnode(13);

newnode->right->right = createnode(55);

/* Sample Tree 1:

* 25

* / \

* 27 19

* / \ / \

* 17 91 13 55

*/

printf(“Sum of nodes in tree 1 = %d”, sumofnodes(newnode));

printf(“\n”);

/* Creating second Tree. */

struct node *node = createnode(1);

node->right = createnode(2);

node->right->right = createnode(3);

node->right->right->right = createnode(4);

node->right->right->right->right = createnode(5);

/* Sample Tree 2: Right Skewed Tree (Unbalanced).

* 1

* \

* 2

* \

* 3

* \

* 4

* \

* 5

*/

printf(“Sum of nodes in tree 2 = %d”, sumofnodes(node));

printf(“\n”);

/* Creating third Tree. */

struct node *root = createnode(15);

/* Sample Tree 3: Tree having just one root node.

* 15

*/

printf(“Sum of nodes in tree 3 = %d”, sumofnodes(root));

printf(“\n”);

return 0;

}

Program Explanation

Program contains two important functions.

1. createnode(key);

This function helps to create a new node by allocating it a memory dynamically. It has just one parameter which is “key” which assigns value to the node thereby creating a fresh node having left and right child as “NULL”.

2. sumofnodes(struct node *root);

In this function we have passed root node of a tree as a parameter and traversed each node of the tree. We have first traversed and added the values(info) of nodes in left subtree and stored it in sum variable then we have traversed the right subtree and added the values(info) of the nodes. Once we traverse the whole tree and add the values of all the nodes we return the value contained in variable “sum”.

Runtime Test Cases

Sum of nodes in tree 1 = 247

Sum of nodes in tree 2 = 15

Sum of nodes in tree 3 = 15