For 1000 elements, there will be 10 levels (with the root at level 1). However, the last level will not be completely filled since we have only 1000 nodes.
As you can see, a portion at the last level is empty. Since the largest element cannot have children (due to missing nodes), we note that a completely filled tree would require 1023 nodes.
Now let’s see if node A (shown below) has any children:
We do not have enough nodes; therefore, A will not have a subtree. This is the key observation in this question. Let’s quickly verify why node A doesn’t have children:
Total nodes till the second-last level = \( 2^9 - 1 = 511 \)
Nodes required to fill all levels completely = \( 2^{10} - 1 = 1023 \)
Let’s analyze:
- For 1019 nodes – both A and B have no children.
- For 1020 nodes – A has a left child.
- For 1021 nodes – A has two children.
- For 1022 nodes – B has a left child.
- For 1023 nodes – B has two children.
Since we have only 1000 nodes, both A and B do not have children.
Finding the 3rd Largest Element:
The largest element is at the rightmost node of the 9th level. The 2nd largest element is at the rightmost node of the 8th level — because to find something larger, you’d have to go further right, but there’s only one element to the right.
Now, the 3rd largest element will be the element just before the 2nd largest in inorder traversal — i.e., the inorder predecessor of the 2nd largest node.
Hence, the 3rd largest element corresponds to node A in the diagram above.
Let’s find the index of the 2nd largest node first (assuming 1-based indexing):
\( 1 \rightarrow 3 \rightarrow 7 \rightarrow 15 \rightarrow 31 \rightarrow 63 \rightarrow 127 \rightarrow 255 \)
Therefore, the index of the 2nd largest = 255.
Now, since the 3rd largest (node A) is its inorder predecessor, it will be at index \( 510 \) (in 1-based indexing).
Since the problem states that indices start from 0, the final answer is:
\( \boxed{509} \)
Therefore, the 3rd largest element of the BST is stored at index 509.
