0426MediumArrayStackTreeBinary Search TreeMonotonic StackBinary Tree

Rebuild the Accession Index

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Trains the technique from

LeetCode 1008Construct Binary Search Tree from Preorder Traversal

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A museum files its specimens in a branching index. Each card in the index carries one accession number and hangs at most two cards below it, one on the low side and one on the high side. The filing rule holds everywhere in the index: every number hanging anywhere below a card on its low side is smaller than that card's number, and every number hanging anywhere below it on its high side is larger. All the numbers are different.

A curator read the index out into entries in card-first order: she wrote down a card's number, then the whole of what hangs on its low side, then the whole of what hangs on its high side, applying that same order at every card she reached. entries[0] is therefore the topmost card.

Rebuild the index and return it as a band listing, one band of the index at a time from the top down, left to right within a band. Slot 0 holds the topmost card. Each card that appears in the listing takes the next two unclaimed slots for what hangs below it, the low side first and the high side second. A slot holding null means nothing hangs there, and such a slot claims no slots of its own. Slots that would hold null at the very end of the listing must be left off.

Exactly one index matches a given card-first reading, so the listing is unique.

Examples

Example 1

Input: entries = [9, 4, 2, 6, 15, 20]
Output: [9, 4, 15, 2, 6, null, 20]

The topmost card is 9. The reading covers all of its low side next, namely 4 with 2 below on the low side and 6 below on the high side, and only then its high side, namely 15 with 20 hanging high of it.

Example 2

Input: entries = [3, 2, 1]
Output: [3, 2, null, 1]

Each number read is smaller than the one before it, so every card hangs on the low side of the card just read, and the index is a single chain down the low side.

Example 3

Input: entries = [1]
Output: [1]

One card and nothing hanging below it, so the listing has a single slot.

Example 4

Input: entries = [50, 30, 20, 40, 70, 60, 80]
Output: [50, 30, 70, 20, 40, 60, 80]

The index comes out full to three bands: 50 on top, then 30 and 70, then 20, 40, 60 and 80.

Example 5

Input: entries = [4, 2, 1, 6, 5, 7]
Output: [4, 2, 6, 1, null, 5, 7]

Card 2 has 1 on its low side and nothing on its high side, so the slot for that high side sits in the middle of the listing holding `null`.

Constraints

1 <= entries.length <= 100
1 <= entries[i] <= 1000
All the values of entries are unique.
entries is the card-first reading of some index obeying the filing rule.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def bst_from_preorder(entries: list[int]) -> list[int | None]:

Java

public Integer[] bstFromPreorder(int[] entries)

The intended approach

Time O(n) for n entries: every entry is pushed and popped at most once, and the band walk visits each card once and each empty slot once, and there are at most n + 1 empty slots.Space O(n) for the low and high tables, the stack and the queue, plus the returned listing, whose length is under 2n.

In a card-first reading, an entry hangs on the low side of the card just before it whenever it is smaller than that card, and otherwise it hangs on the high side of the deepest earlier card it is bigger than. Keep those still-open cards in a stack, which stays in falling order from bottom to top; for each entry, pop every card it is bigger than and hang it on the high side of the last one popped, or on the low side of whatever is left on top if it popped nothing. Each entry is pushed once and popped at most once. A band-by-band walk of the finished index then emits the listing, queueing two slots for every card written and none for an empty slot, and the trailing empty slots are trimmed.

Pythonentry point: bst_from_preorder

def bst_from_preorder(entries):
    # A record entry is a right branch exactly when it overtakes some ancestor.
    # Keep the chain of ancestors still open in a stack whose numbers fall from
    # the bottom up; the deepest ancestor the new entry overtakes is the one it
    # hangs to the right of, and if it overtakes nobody it hangs to the left of
    # the entry on top.
    size = len(entries)
    left = [-1] * size
    right = [-1] * size
    open_chain = [0]
    for pos in range(1, size):
        overtaken = -1
        while open_chain and entries[open_chain[-1]] < entries[pos]:
            overtaken = open_chain.pop()
        if overtaken == -1:
            left[open_chain[-1]] = pos
        else:
            right[overtaken] = pos
        open_chain.append(pos)

    # band-by-band listing: an empty slot claims no slots of its own
    listing = []
    queue = [0]
    head = 0
    while head < len(queue):
        node = queue[head]
        head += 1
        if node == -1:
            listing.append(None)
            continue
        listing.append(entries[node])
        queue.append(left[node])
        queue.append(right[node])
    while listing and listing[-1] is None:
        listing.pop()
    return listing

The obvious version

Split by scanning: the first number of a stretch is its card, the numbers after it that are smaller form the low side, and everything from the first larger number onward forms the high side. Recurse on those two stretches to build a nested structure, then read it band by band into a listing. O(n^2).

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_bst_from_preorder(entries):
    size = len(entries)

    def shape(lo, hi):
        if lo >= hi:
            return None
        root = entries[lo]
        split = hi
        for probe in range(lo + 1, hi):
            if entries[probe] > root:
                split = probe
                break
        return [root, shape(lo + 1, split), shape(split, hi)]

    tree = shape(0, size)
    if tree is None:
        return []
    listing = []
    row = [tree]
    while row:
        nxt = []
        for spot in row:
            if spot is None:
                listing.append(None)
            else:
                listing.append(spot[0])
                nxt.append(spot[1])
                nxt.append(spot[2])
        row = nxt
    while listing and listing[-1] is None:
        listing.pop()
    return listing

How this problem was checked

Reference solution executed against 16 test cases — 16 passed
5 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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