Sum of Each Branch in a Binary Search Tree - c++

Question

My assignment is to find the sum of all nodes on each branch in a binary search tree using recursion, and compare them to a user input value. If the user input value matches a sum of one of the branches, the function should return true.
In other words, the sum of 32+24+21+14=91. The sum of 32+24+28+25=109. The sum of 32+24+28+31=115 etc. I have tried many different methods, but cant seem to figure out how to traverse each branch accurately. So far I have only been able to traverse and find the sum of the left-most branch.
I am using the method of subtracting each node from the user input value. If the value reaches 0 at a Leaf-node, then clearly the user-input matches the node-sum of that branch on the tree.
The particular points of difficulty for me are when the branch diverges, such as at the node [24] and [28]. I clearly am getting something very simple wrong, but I cant figure it out.
Below is the condensed code I've written so far, in the form of two companion methods (also required for the assignment).
public:
bool findBranchSum1(int value) throw (InvalidTreeArgument) {
if (root == nullptr)
throw InvalidTreeArgument();
return(findBranchSum(root, value));
}
private:
bool findBranchSum(NodePtr node, int value) throw (InvalidTreeArgument)
{
bool result = false;
if (root == nullptr)
throw InvalidTreeArgument();
value -= node->getElement(); //subtract current node from user-input value.
cout << "Current Value = " << value << endl; //help track value changes
if (node->getLeftSide() == nullptr && node->getRightSide() == nullptr)
{
if (value == 0)
{
result = true;
return(true);
}
else
return(false);
}
else
{
if (node->getLeftSide() != nullptr)
{
node = node->getLeftSide(); //advance to next Left node
result = findBranchSum(node, value); //recursive call using new node
}
if (node->getRightSide() != nullptr)
{
node = node->getRightSide(); //advance to next Right node
result = findBranchSum(node, value); //recursive call using new node
}
return(result);
}
}
What am I doing wrong, and how can I fix my code to find the sum of each branch on the tree? Thank you in advance. I apologize for any errors in my format, or missing information.

Answer 1

This is wrong:
if (node->getLeftSide() != nullptr)
{
node = node->getLeftSide(); //advance to next Left node
result = findBranchSum(node, value); //recursive call using new node
}
if (node->getRightSide() != nullptr)
{
node = node->getRightSide(); //advance to next Right node
result = findBranchSum(node, value); //recursive call using new node
}
because you move to the left and then to the right branch of the left (node is changed by your assignment), if it exists! Change to:
if (node->getLeftSide() != nullptr)
{
result = findBranchSum(node->getLeftSide(), value);
}
if (node->getRightSide() != nullptr)
{
result = findBranchSum(node->getRightSide(), value);
}
Your return value management is also broken, change it to:
if (node->getLeftSide() != nullptr)
{
result = findBranchSum(node->getLeftSide(), value);
}
if (!result && node->getRightSide() != nullptr) // cut exploration if previous was correct...
{
result = findBranchSum(node->getRightSide(), value);
}
return result;
if you need to stop at the first correct branch.

Answer 2

I might try something like the following.
bool IsLeaf(Node const * node) {
return node && !node->left && !node->right;
}
bool CheckPathSum(Node const * node, int const target, int const sum_so_far) {
if (!node) return false;
int const sum = sum_so_far + node->element;
if IsLeaf(node) && (sum == target) return true;
return CheckPathSum(node->left, target, sum) ||
CheckPathSum(node->right, target, sum);
}
Call as
CheckPathSum(root, target, 0);

Answer 3

In Java, i tried this-
private static void branchSumsUtil(TreeNode root, List<Integer> sumArray, int runningSum) {
if (root == null){
return;
}
int newRunningSum = runningSum + root.key;
if (root.left == null && root.right == null){
sumArray.add(newRunningSum);
}
branchSumsUtil(root.left, sumArray, newRunningSum);
branchSumsUtil(root.right, sumArray, newRunningSum);
}