0751MediumTreeBreadth-First SearchBinary Tree

Reading A Binary Tree Deepest Level First

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Trains the technique from

LeetCode 107Binary Tree Level Order Traversal II

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 binary tree arrives as nodes, a flat list written down by walking the tree one level at a time from left to right.

nodes[0] holds the root's value.
After that the list carries two slots for every node already written down, taken in the same order those nodes were written: first that node's left child, then its right child.
A slot holding null means the child is absent. An absent child is not a node, so it claims no slots of its own further down the list.
Slots at the very end of the list may be left off when every one of them would be null.
An empty list means the tree has no nodes.

So [4, null, 6, 2] describes a root of 4 whose left child is absent, whose right child is 6, and whose grandchild 2 is the left child of 6.

Return the values of the tree grouped by level, with the deepest level first and the root's level last. Within each group the values run left to right. For a tree with no nodes return an empty list.

Examples

Example 1

Input: nodes = [5, null, 8, 1, 4]
Output: [[1, 4], [8], [5]]

The root is `5`, its left slot is `null` and its right child is `8`. The next two slots belong to `8`, giving it a left child `1` and a right child `4`. Taking the levels from the deepest upward gives `[1, 4]`, then `[8]`, then `[5]`.

Example 2

Input: nodes = [6, 2, 9, null, 7]
Output: [[7], [2, 9], [6]]

The root `6` has children `2` and `9`. The four slots of the level below belong to `2` and then `9`, and the list supplies only two of them, so `2` has no left child and a right child `7` while `9` has none. The deepest level is `[7]`, above it `[2, 9]`, and the root's level last.

Example 3

Input: nodes = [42]
Output: [[42]]

The tree is a single root with both child slots left off the end of the list, so there is one level to report.

Constraints

0 <= nodes.length <= 4001
-1000 <= nodes[i] <= 1000
The tree holds between 0 and 2000 nodes.
Every entry of nodes is either an integer or null, and nodes[0] is never null.

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 levels_deepest_first(nodes: list) -> list[list[int]]:

Java

public List<List<Integer>> levelsDeepestFirst(Integer[] nodes)

The intended approach

Time O(k) for a tree of k nodes, plus O(len(nodes)) to scan the flat list.Space O(k) for the rebuilt tree, the queues and the returned levels.

Two queue-driven passes. The first rebuilds the tree: hold the nodes discovered so far in a list, take them in discovery order, and for each one consume the next two entries of the flat list as its left and right slots, recording a child and adding it to the discovery queue only when the slot is not `null`. Because the flat list was written in that same order, this pairs every slot with the right parent even when absent children shift things along. The second pass holds the current level as a list of node positions, records their values, then builds the next level from their present children, until a level comes out empty. Reversing the collected levels puts the deepest one first. Both passes are loops over explicit queues, so a 2000-node chain does not overflow anything.

Pythonentry point: levels_deepest_first

def levels_deepest_first(nodes):
    # rebuild the tree from the flat list with a queue, then sweep it level by
    # level with a second queue and reverse the collected levels.
    # O(k) time and O(k) space for k nodes. No recursion, so a 2000-node chain
    # is safe.
    if not nodes or nodes[0] is None:
        return []

    val = [nodes[0]]
    kids = [[-1, -1]]
    order = [0]
    head = 0
    pos = 1
    while head < len(order) and pos < len(nodes):
        cur = order[head]
        head += 1
        for side in (0, 1):
            if pos >= len(nodes):
                break
            entry = nodes[pos]
            pos += 1
            if entry is None:
                continue
            val.append(entry)
            kids.append([-1, -1])
            child = len(val) - 1
            kids[cur][side] = child
            order.append(child)

    levels = []
    frontier = [0]
    while frontier:
        levels.append([val[i] for i in frontier])
        nxt = []
        for i in frontier:
            for c in kids[i]:
                if c != -1:
                    nxt.append(c)
        frontier = nxt
    levels.reverse()
    return levels

The obvious version

Slice the flat list into levels without ever building a tree: the first level takes one slot, and each later level takes two slots for every value that was present on the level before. Keep the non-null values of each slice as that level, stop when a slice has none, and reverse.

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_levels_deepest_first(nodes):
    """Never builds a tree. Reads the flat list one level at a time: the first
    level occupies one slot, and each later level occupies two slots per node
    that was present on the level before it."""
    if not nodes or nodes[0] is None:
        return []
    levels = []
    pos = 0
    slots = 1
    while slots > 0 and pos < len(nodes):
        chunk = nodes[pos:pos + slots]
        pos += slots
        present = [v for v in chunk if v is not None]
        if not present:
            break
        levels.append(present)
        slots = 2 * len(present)
    levels.reverse()
    return levels

How this problem was checked

Reference solution executed against 14 test cases — 14 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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