C++: Sum of all node values of a binary tree - c++

Question

I'm preparing for a job interview. I was stuck at one of the binary tree questions:
How can we calculate the sum of the values present in all the nodes of a binary tree?

Answer 1

The elegant recursive solution (in pseudo-code):
def sum (node):
if node == NULL:
return 0
return node->value + sum (node->left) + sum (node->right)
then just use:
total = sum (root)
This correctly handles the case of a NULL root node.
And if you want to see it in action in C++, here's some code using that algorithm. First, the structure for a node and the sum function:
#include <iostream>
typedef struct sNode {
int value;
struct sNode *left;
struct sNode *right;
} tNode;
int sum (tNode *node) {
if (node == 0) return 0;
return node->value + sum (node->left) + sum (node->right);
}
Then the code below is a test harness code for inserting nodes:
static tNode *addNode (tNode *parent, char leftRight, int value) {
tNode *node = new tNode();
node->value = value;
node->left = 0;
node->right = 0;
if (parent != 0) {
if (leftRight == 'L') {
parent->left = node;
} else {
parent->right = node;
}
}
return node;
}
And, finally, the main function for constructing the following tree, one that covers all of the valid possibilities (empty node, node with two children, node with no children, node with one right child and node with one left child):
10
/ \
7 20
/ \
3 99
\
4
\
6
The code to construct that tree and report the sum at various points is shown here:
int main (void) {
// Empty tree first.
tNode *root = 0;
std::cout << sum (root) << '\n';
// Then a tree with single node (10).
root = addNode (0, ' ', 10);
std::cout << sum (root) << '\n';
// Then one with two subnodes (10, 7, 20).
addNode (root,'L',7);
addNode (root,'R',20);
std::cout << sum (root) << '\n';
// Then, finally, the full tree as per above.
addNode (root->left,'L',3);
addNode (root->left->left,'R',4);
addNode (root->left->left->right,'R',6);
addNode (root->right,'R',99);
std::cout << sum (root) << '\n';
return 0;
}
This outputs (the correct):
0
10
37
149

Answer 2

Traverse the tree in any order (pre, post, in). Instead of printing the node calculate the total.
void sum(Node* root, int& total)
{
if(root == NULL)
{
return;
}
sum(root->left, total);
total = total + root->value;
sum(root->right, total);
}
int main()
{
int total =0;
sum(root,total);
cout << total;
}

Answer 3

The same way you search the tree, or display each node, or any other tree-wide operation: visit the current node, visit the left sub-tree (recursively), and visit the right sub-tree (recursively).
Essentially, something like this:
int TreeNode::calculateSum() const
{
int sum = this->value;
if (this->left != NULL) sum += this->left ->calculateSum();
if (this->right != NULL) sum += this->right->calculateSum();
return sum;
}
Because of the if checks the recursion will eventually bottom out when it reaches nodes with no left or right children (leaf nodes).

Answer 4

While the STL has more complex and concise mechanisms for doing this, it's a very fast rode to productivity just to learn to use a manual loop over a container, something like:
Tree::value_type total = Tree::value_type();
for (Tree::const_iterator i = tree.begin(); i != tree.end(); ++i)
total += *i;
This assumes your binary tree is a STL::map, or if not you'll provide an iterator concept for your own implementation....

Answer 5

Use one of the Tree Traversal techniques(In-order, Pre-order, Post-order) to visit each node and store the sum in a variable.
You can find more details on tree traversal in this wiki

Answer 6

50
/ \
30 70
/ \ / \
20 40 60 80
returns:350
int Sum(struct node *root)
{
if(root->left == NULL && root->right== NULL)
return root->key;
int lvalue,rvalue;
lvalue=Sum(root->left);
rvalue=Sum(root->right);
return root->key+lvalue+rvalue;
}

Answer 7

public int sum(Node root){
if(root==null){
return 0;
}
if(root.left == null && root.right==null){
return root.key;
}
return sum(root.left)+sum(root.right)+root.key;
}