0546MediumTreeDepth-First SearchBinary Tree

Outline of a Printed Tree Diagram

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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 screen reader has to announce the labels printed around the outside of a binary tree diagram, tracing anticlockwise from the top.

The diagram arrives as root, a level-order list. Its first entry is the top node's label. After that, the list gives the two children of each non-null node already listed, in the order those nodes appear, left child first, using null where a child is absent. Trailing null entries are left off.

Call a node a tip when it has no children. Two chains matter:

The left rail is empty when the top node has no left child. Otherwise begin at the top node's left child and repeatedly move to the current node's left child, or to its right child when there is no left child, stopping the moment a tip is reached. The rail holds the labels of the nodes stepped through before that tip, so no tip is ever on the rail.
The right rail is built the same way from the top node's right child, except that at each step the right child is taken when it exists and the left child only otherwise.

Return a list holding, in this order: the top node's label; the left rail from top to bottom; every tip in the order met by a walk that always goes into a node's left child before its right child; and the right rail from bottom to top. A diagram of a single node returns a list holding that one label.

Examples

Example 1

Input: root = [8, 3, 10, null, 6, 11, 20, 4, 7, 12]
Output: [8, 3, 6, 4, 7, 12, 20, 10]

The top label 8 comes first. The left rail starts at 3, which has no left child, so it steps right to 6; 6 has the left child 4, a tip, so the rail is 3 then 6. The tips read left to right are 4, 7, 12 and 20. The right rail starts at 10 and steps right to the tip 20, so it holds 10 alone.

Example 2

Input: root = [5, 4, null, 3, null, 2]
Output: [5, 4, 3, 2]

This diagram leans entirely left. The left rail runs 4 then 3 and stops at the tip 2, which is the only tip. The top node has no right child, so the right rail is empty, giving 5, 4, 3, 2.

Example 3

Input: root = [9]
Output: [9]

A single node is both the top of the diagram and a tip, and the rule for a one-node diagram lists its label once.

Constraints

The number of nodes in the tree is in the range [1, 10^4].
-1000 <= Node.val <= 1000
root is a level-order list as described, so root[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 boundary_of_binary_tree(root: list[int | None]) -> list[int]:

Java

public List<Integer> boundaryOfBinaryTree(Integer[] root)

The intended approach

Time O(n): the rebuild, each rail and the tip walk all touch a node a constant number of times.Space O(n) for the rebuilt tree, the stack and the returned list.

Rebuild the tree from the level-order list with a queue, then treat the outline as three independent pieces. The left rail is a plain downward loop that prefers the left child and halts at the first tip; the right rail is the mirror of that loop, collected downward and then reversed so it reads bottom to top. The tips are gathered by a stack-driven walk that pushes the right child before the left child, so children come off in left-to-right order. No node can land on both a rail and the tip list, because rails stop before tips, and the two rails live in different subtrees, so nothing needs de-duplicating. A one-node diagram is answered before any of that, since its only node is both the top and a tip. Every walk is a loop rather than a call chain, so a 10^4-node chain is fine.

Pythonentry point: boundary_of_binary_tree


def _grow(vals):
    if not vals or vals[0] is None:
        return None
    top = {"v": vals[0], "l": None, "r": None}
    order = [top]
    at = 1
    head = 0
    while head < len(order) and at < len(vals):
        cur = order[head]
        head += 1
        if at < len(vals):
            v = vals[at]
            at += 1
            if v is not None:
                cur["l"] = {"v": v, "l": None, "r": None}
                order.append(cur["l"])
        if at < len(vals):
            v = vals[at]
            at += 1
            if v is not None:
                cur["r"] = {"v": v, "l": None, "r": None}
                order.append(cur["r"])
    return top


def boundary_of_binary_tree(root):
    # O(n) time. Three separate walks, none of them recursive, so a chain of
    # 10^4 nodes is safe: slide down the left rail, gather the tips with an
    # explicit stack that pushes the right child first so the left child comes
    # off next, then slide down the right rail and reverse it.
    top = _grow(root)
    if top is None:
        return []

    def tip(node):
        return node["l"] is None and node["r"] is None

    if tip(top):
        return [top["v"]]

    left = []
    cur = top["l"]
    while cur is not None and not tip(cur):
        left.append(cur["v"])
        cur = cur["l"] if cur["l"] is not None else cur["r"]

    tips = []
    stack = [top]
    while stack:
        node = stack.pop()
        if tip(node):
            tips.append(node["v"])
            continue
        if node["r"] is not None:
            stack.append(node["r"])
        if node["l"] is not None:
            stack.append(node["l"])

    right = []
    cur = top["r"]
    while cur is not None and not tip(cur):
        right.append(cur["v"])
        cur = cur["r"] if cur["r"] is not None else cur["l"]
    right.reverse()

    return [top["v"]] + left + tips + right

The obvious version

Rebuild the tree, then catalogue every node together with the sequence of left and right turns that reaches it from the top. Sorting that catalogue by turn sequence, with a left turn ordered before a right turn, puts the nodes in drawing order, so filtering it down to childless nodes gives the tips. The two rails are collected by a recursive descent that appends the current label and steps on until it hits a tip. Concatenate the pieces.

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 _grow(vals):
    if not vals or vals[0] is None:
        return None
    top = {"v": vals[0], "l": None, "r": None}
    order = [top]
    at = 1
    head = 0
    while head < len(order) and at < len(vals):
        cur = order[head]
        head += 1
        if at < len(vals):
            v = vals[at]
            at += 1
            if v is not None:
                cur["l"] = {"v": v, "l": None, "r": None}
                order.append(cur["l"])
        if at < len(vals):
            v = vals[at]
            at += 1
            if v is not None:
                cur["r"] = {"v": v, "l": None, "r": None}
                order.append(cur["r"])
    return top


def bf_boundary_of_binary_tree(root):
    top = _grow(root)
    if top is None:
        return []

    def tip(node):
        return node["l"] is None and node["r"] is None

    if tip(top):
        return [top["v"]]

    catalogue = []

    def visit(node, path):
        catalogue.append((path, node))
        if node["l"] is not None:
            visit(node["l"], path + (0,))
        if node["r"] is not None:
            visit(node["r"], path + (1,))

    visit(top, ())
    catalogue.sort(key=lambda pair: pair[0])
    tips = [node["v"] for path, node in catalogue if tip(node)]

    def rail(node, first, second, acc):
        if node is None or tip(node):
            return acc
        acc.append(node["v"])
        step = node[first]
        if step is None:
            step = node[second]
        return rail(step, first, second, acc)

    left = rail(top["l"], "l", "r", [])
    right = rail(top["r"], "r", "l", [])
    right.reverse()
    return [top["v"]] + left + tips + right

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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