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Asked: May 10, 20262026-05-10T17:50:08+00:00 2026-05-10T17:50:08+00:00

This may be a simple fix – but I’m trying to sum together all

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This may be a simple fix – but I’m trying to sum together all the nodes (Size property from the Node class) on the binary search tree. Below in my BST class I have the following so far, but it returns 0:

    private long sum(Node<T> thisNode)     {         if (thisNode.Left == null && thisNode.Right == null)             return 0;         if (node.Right == null)             return sum(thisNode.Left);         if (node.Left == null)              return sum(thisNode.Right);           return sum(thisNode.Left) + sum(thisNode.Right);     } 

Within my Node class I have Data which stores Size and Name in their given properties. I’m just trying to sum the entire size. Any suggestions or ideas?

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  1. 2026-05-10T17:50:09+00:00Added an answer on May 10, 2026 at 5:50 pm

    It’s because you’re returning zero when you reach a leaf node. You should be returning the size stored in that leaf node.

    In addition, if your non-leaf nodes also have a size, you’ll need to process them as well thus:

    private long sum(Node<T> thisNode) {     if (thisNode.Left == null && thisNode.Right == null)         return thisNode.Size;     if (node.Right == null)         return thisNode.Size + sum(thisNode.Left);     if (node.Left == null)          return thisNode.Size + sum(thisNode.Right);     return thisNode.Size + sum(thisNode.Left) + sum(thisNode.Right); } 

    If your non-leaf nodes don’t have size, use:

    private long sum(Node<T> thisNode) {     if (thisNode.Left == null && thisNode.Right == null)         return thisNode.Size;     if (node.Right == null)         return sum(thisNode.Left);     if (node.Left == null)          return sum(thisNode.Right);     return sum(thisNode.Left) + sum(thisNode.Right); } 

    A more elegant version of the first one is:

    private long sum(Node<T> thisNode) {     if (thisNode == null)         return 0;     return thisNode.Size + sum(thisNode.Left) + sum(thisNode.Right); } 
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