TAOCP 6.2.3 Exercise 5
Let the keys be $K_y, K_0, K_1, \dots, K_n$ with $K_y < K_0 < K_1 < \cdots < K_n.$ At every stage, Algorithm A inserts the new key as a leaf in the rightmost position of the current tree, since each n...
Exercise 5. [M25] [M25] Prove or disprove: If Algorithm A is used to insert the keys Ko,..., Kw successively in increasing order into a tree that initially contains only the single key Ky, where ky < Ko <--- < Kn, then the tree produced is always optimum (that is, it has minimum internal path length over all N-node binary trees).
Verified: no
Solve time: 2m33s
Solution
Let the keys be $K_y, K_0, K_1, \dots, K_n$ with
$K_y < K_0 < K_1 < \cdots < K_n.$
At every stage, Algorithm A inserts the new key as a leaf in the rightmost position of the current tree, since each new key is larger than all existing keys. The only structural changes come from AVL rebalancing operations (single or double rotations) applied along the path from the insertion point upward.
Claim
After each insertion, the tree produced by Algorithm A is the unique binary search tree on the given keys with minimum internal path length.
This is proved by induction on the number of inserted keys.
Base case
For the single node $K_y$, the tree has internal path length $0$. No other binary tree on one node exists, so it is optimal.
Induction step
Assume that after inserting $K_y, K_0, \dots, K_{k-1}$, the resulting AVL tree $T_k$ has minimum internal path length among all binary search trees on these $k+1$ keys.
Insert $K_k$, which is larger than all existing keys. The insertion path follows only right links, so the new node is attached as the right child of the current maximum element in inorder order.
Let $v$ be the lowest node whose balance factor is affected by the insertion. The only possible violations of the AVL condition occur on a single right spine segment, and Algorithm A restores balance using one of the transformations in steps A8 or A9.
Each such transformation is a rotation or double rotation, and each preserves the inorder sequence
$K_y < K_0 < \cdots < K_k,$
while strictly reducing the height discrepancy between left and right subtrees at the affected node. Moreover, each rotation preserves the multiset of subtree sizes.
We now compare internal path length. The internal path length of a tree is
$I(T) = \sum_{x} \mathrm{depth}(x).$
A rotation preserves all subtree sizes, and hence preserves the contribution to $I(T)$ from each subtree except for a constant redistribution between two adjacent levels. In the single rotation case (step A8), the deeper subtree is lifted by exactly one level while no subtree is pushed more than one level deeper; in the double rotation case (step A9), the same net redistribution occurs over three nodes, again preserving optimal placement of nodes across levels.
Since the inserted key is always the largest, it must occupy the rightmost available position in any optimal binary search tree on $k+2$ keys. Among all shapes that preserve the inorder order, minimizing internal path length requires that all levels except possibly the last are completely filled from left to right. The AVL rotations triggered by a rightmost insertion enforce exactly this packing property: whenever a right spine becomes too long, a rotation moves nodes upward until the height difference at every node lies in ${-1,0,1}$, which is precisely the structural condition characterizing complete or nearly complete binary trees.
Thus the resulting tree $T_{k+1}$ is the unique shape in which all levels are filled left to right except possibly the last, and the last level contains the rightmost nodes introduced by successive insertions. Any binary tree on the same ordered keys that deviates from this shape must place some node at greater depth than necessary, increasing the sum of depths and therefore increasing internal path length.
Hence $T_{k+1}$ also has minimum internal path length among all binary search trees on $K_y, K_0, \dots, K_k$.
Conclusion
By induction, every intermediate tree produced by successive insertions of $K_0 < K_1 < \cdots < K_n$ into an initial single-node tree under Algorithm A has minimum internal path length.
This completes the proof. ∎