Consider a binary search tree (BST) with $n$ leaf nodes ($n > 0$). Given any node $V$, the key present in the node is denoted as $Val(V)$. All the keys present in the given BST are distinct. The keys belong to the set of real numbers. For a node $V$, let $Suc(V)$ denote the node that is its inorder successor. If a node $V$ does not have an inorder successor, then $Suc(V)$ is $NULL$. As there are no duplicates, if $Suc(V)$ is not $NULL$, then $Val(V) < Val(Suc(V))$. Corresponding to every leaf node $L_i$ that has a non-$NULL$ $Suc(L_i)$, a new key $k_i$ with the following property is to be inserted into the BST. $$Val(L_i) < k_i < Val(Suc(L_i))$$ Let $K$ represent the list of all such new keys to be inserted into the BST. Which of the following statements is/are true?

Let us analyze each statement based on the properties of a Binary Search Tree (BST):

1. Option A: Since all keys in the BST are distinct, each leaf node $L_i$ has its successor $Suc(L_i)$ strictly greater than $Val(L_i)$, with no other existing tree keys between $Val(L_i)$ and $Val(Suc(L_i))$. Therefore, for two distinct leaf nodes $L_i$ and $L_j$ with $Val(L_i) < Val(L_j)$, the intervals $(Val(L_i), Val(Suc(L_i)))$ and $(Val(L_j), Val(Suc(L_j)))$ are completely disjoint. Since each $k_i$ belongs to its unique open interval, no two elements in $K$ can be equal. Hence, $K$ cannot contain any duplicates. (Statement A is true)

2. Option B: Consider a BST with only $1$ node (the root itself), so $n = 1$. The root is the only leaf node and also the maximum element in the tree, so its inorder successor is NULL. Thus, no leaf node has a non-NULL successor, resulting in $K$ being empty ($|K| = 0$). Hence, $K$ does not necessarily have at least one element. (Statement B is false)

3. Option C: For each leaf node $L_i$, any key $k_i$ satisfying $Val(L_i) < k_i < Val(Suc(L_i))$ will follow the BST insertion path down to $L_i$ and will be attached directly as the right child of $L_i$ (since $L_i$ is a leaf and has no children). Because each inserted key attaches as an immediate child of an existing leaf node, the depth of any leaf increases by at most $1$. Consequently, the total height of the BST can increase by at most $1$. (Statement C is true)

4. Option D: If a BST has total nodes $N > n$, inserting at most $n$ keys yields at most $N + n$ nodes, which is strictly less than $2N$. Thus, the number of nodes will not necessarily double. (Statement D is false)

Therefore, the correct options are A and C.