0283MediumTreeBinary Search TreeBinary Tree

Retire a Locker from the Sorted Index

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Trains the technique from

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 depot keeps its locker codes in a binary search tree: every code in a node's left subtree is smaller than that node's code and every code in its right subtree is larger, and no code is stored twice.

The tree arrives as index, a level-by-level listing. index[0] is the root. After that, every node already listed contributes two entries in turn, its left child then its right child, with null where that child is missing. A null contributes nothing further, and null entries at the very end are left off. An empty listing means the tree is empty.

The locker with code code is being retired. Take its node out of the tree and return the resulting tree in the same level-by-level listing, again with trailing null entries left off. When no node holds code, hand back the tree as it arrived.

So that exactly one listing is correct, removal follows these rules:

a node with no children is dropped;
a node with a child on one side only is replaced by that child, whole subtree included;
a node with children on both sides stays where it is and takes on the smallest code held in its right subtree, and the node that held that smallest code is then taken out of the right subtree under these same rules.

What comes back is still a binary search tree.

Examples

Example 1

Input: index = [50, 30, 70, 20, 40, 60, 80], code = 50
Output: [60, 30, 70, 20, 40, null, 80]

The root holds 50 and has children on both sides. The smallest code in its right subtree is 60, so the root takes on 60, and the node that held 60 has no children of its own and is dropped from under 70.

Example 2

Input: index = [50, 30, 70, 20, 40, 60, 80], code = 20
Output: [50, 30, 70, null, 40, 60, 80]

The node holding 20 has no children, so it is dropped and every other node stays where it was.

Example 3

Input: index = [50, 30, 70, 20, 40, 60, 80], code = 99
Output: [50, 30, 70, 20, 40, 60, 80]

No node holds 99, so the listing comes back exactly as it arrived.

Constraints

The number of nodes in the tree is in the range [0, 10^4].
-10^5 <= node code <= 10^5
Each node has a unique code.
index is a valid binary search tree.
-10^5 <= code <= 10^5

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 retire_locker(index: list[int | None], code: int) -> list[int | None]:

Java

public Integer[] retireLocker(Integer[] index, int code)

The intended approach

Time O(n)Space O(n)

Parse the listing into value and child index tables, then walk down from the root comparing against each node's code and holding on to the parent, so the node and the link pointing at it are both in hand. A node with a child on one side only, or none, is unhooked by pointing its parent at that child. A node with children on both sides keeps its slot and receives the code of the leftmost node in its right subtree, and that leftmost node is unhooked from whichever parent holds it, which is the node itself when the right child has no left child. Finally the tree is written back out level by level with trailing gaps trimmed.

Pythonentry point: retire_locker

def retire_locker(index, code):
    # Parse the listing into value and child tables, then walk down from the root
    # using the ordering to find the node and remember its parent, so nothing
    # recurses even on a chain of 10000 codes.  A node with children on both sides
    # keeps its slot and receives the smallest code from its right subtree, which is
    # the leftmost node there; that leftmost node is then spliced out of whichever
    # parent holds it.  Otherwise the node's only child, if any, is hung on the
    # parent in its place.
    # Time: O(n) - O(h) for the descent, plus O(n) to parse and to write the listing.
    # Space: O(n) for the parsed tables and the listing returned.
    n = len(index)
    if n == 0 or index[0] is None:
        return []
    val = list(index)
    left = [-1] * n
    right = [-1] * n
    order = [0]
    head = 0
    i = 1
    while head < len(order) and i < n:
        node = order[head]
        head += 1
        if i < n:
            if index[i] is not None:
                left[node] = i
                order.append(i)
            i += 1
        if i < n:
            if index[i] is not None:
                right[node] = i
                order.append(i)
            i += 1
    root = 0
    parent = -1
    cur = root
    while cur >= 0 and val[cur] != code:
        parent = cur
        cur = left[cur] if code < val[cur] else right[cur]
    if cur >= 0:
        if left[cur] >= 0 and right[cur] >= 0:
            holder = cur
            small = right[cur]
            while left[small] >= 0:
                holder = small
                small = left[small]
            val[cur] = val[small]
            if holder == cur:
                right[holder] = right[small]
            else:
                left[holder] = right[small]
        else:
            child = left[cur] if left[cur] >= 0 else right[cur]
            if parent < 0:
                root = child
            elif left[parent] == cur:
                left[parent] = child
            else:
                right[parent] = child
    if root < 0:
        return []
    out = []
    queue = [root]
    h = 0
    while h < len(queue):
        node = queue[h]
        h += 1
        if node < 0:
            out.append(None)
            continue
        out.append(val[node])
        queue.append(left[node])
        queue.append(right[node])
    while out and out[-1] is None:
        out.pop()
    return out

The obvious version

Rebuild the listing as linked dictionaries, then remove the code with a recursive function that returns the new subtree at every level, re-hanging each returned subtree on its parent, and serialise whatever root comes back.

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_retire_locker(index, code):
    if not index or index[0] is None:
        return []
    boxes = [None if v is None else {"v": v, "l": None, "r": None} for v in index]
    pend = [boxes[0]]
    at = 0
    idx = 1
    while at < len(pend) and idx < len(index):
        cur = pend[at]
        at += 1
        for side in ("l", "r"):
            if idx >= len(index):
                break
            ch = boxes[idx]
            idx += 1
            if ch is not None:
                cur[side] = ch
                pend.append(ch)

    def drop(node, key):
        if node is None:
            return None
        if key < node["v"]:
            node["l"] = drop(node["l"], key)
            return node
        if key > node["v"]:
            node["r"] = drop(node["r"], key)
            return node
        if node["l"] is None:
            return node["r"]
        if node["r"] is None:
            return node["l"]
        small = node["r"]
        while small["l"] is not None:
            small = small["l"]
        node["v"] = small["v"]
        node["r"] = drop(node["r"], small["v"])
        return node

    root = drop(boxes[0], code)
    if root is None:
        return []
    out = []
    q = [root]
    h = 0
    while h < len(q):
        nd = q[h]
        h += 1
        if nd is None:
            out.append(None)
            continue
        out.append(nd["v"])
        q.append(nd["l"])
        q.append(nd["r"])
    while out and out[-1] is None:
        out.pop()
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 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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