0375MediumTreeDepth-First SearchBinary Tree

Total of the Menu Paths

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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 menu tree arrives as the level-order list root.

The first entry is the digit held by the root. After that the entries come level by level, left to right, and every node that is present contributes two entries of its own, one per child slot, with null standing for an empty slot. Absent nodes contribute nothing, and trailing null entries are left off the end. So [7, null, 4] is a root holding 7 whose only child is a right child holding 4.

Every node holds one digit from 0 to 9. Walking from the root down to a leaf spells a number, the root's digit being the most significant and the leaf's digit the least. A leaf is a node with both child slots empty. A walk that starts on a 0 still reads as an ordinary number, so 0 followed by 4 spells 4.

Return the total of the numbers spelled by every walk from the root to a leaf.

Examples

Example 1

Input: root = [6,2,7,4,null,null,3]
Output: 1297

Two walks reach a leaf. They spell 624 and 673, and 624 + 673 is 1297.

Example 2

Input: root = [1,null,2,null,3]
Output: 123

Each node has a single right child, so the one walk to a leaf spells 123.

Example 3

Input: root = [0,4]
Output: 4

The only walk spells 0 then 4, which reads as the number 4.

Example 4

Input: root = [5]
Output: 5

The root has no children, so it is itself a leaf and the only walk spells 5.

Constraints

The number of nodes in the tree is in the range [1, 1000].
0 <= Node.val <= 9
The depth of the tree will not exceed 10.

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 sum_numbers(root: list) -> int:

Java

public int sumNumbers(Integer[] root)

The intended approach

Time O(n)Space O(n) for the child arrays and the stack

First turn the level-order list into two child-index arrays by replaying the queue the encoding was written from: pop a node, consume the next two entries as its child slots, and enqueue the entries that are not null. Then walk the tree from an explicit stack, carrying the number spelled so far and extending it with the current digit as running * 10 + digit. When a node has both slots empty, add its running value to the total.

Pythonentry point: sum_numbers

def _links(root):
    n = len(root)
    left = [None] * n
    right = [None] * n
    order = [0]
    head = 0
    i = 1
    while head < len(order) and i < n:
        cur = order[head]
        head += 1
        if i < n:
            if root[i] is not None:
                left[cur] = i
                order.append(i)
            i += 1
        if i < n:
            if root[i] is not None:
                right[cur] = i
                order.append(i)
            i += 1
    return left, right


def sum_numbers(root):
    # Explicit stack rather than recursion, so nothing depends on the interpreter
    # depth limit. Time O(n), space O(n) for the link arrays and the stack.
    left, right = _links(root)
    total = 0
    stack = [(0, 0)]
    while stack:
        node, running = stack.pop()
        running = running * 10 + root[node]
        l, r = left[node], right[node]
        if l is None and r is None:
            total += running
            continue
        if l is not None:
            stack.append((l, running))
        if r is not None:
            stack.append((r, running))
    return total

The obvious version

Re-index the nodes into the slots of a notional perfect tree, where a node at slot p keeps its children at 2p+1 and 2p+2, then collect the digit string of every root-to-leaf walk and add up the strings read as integers.

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_sum_numbers(root):
    n = len(root)
    slots = {0: root[0]}
    level = [0]
    i = 1
    while level and i < n:
        nxt = []
        for pos in level:
            for side in (1, 2):
                if i >= n:
                    break
                value = root[i]
                i += 1
                if value is not None:
                    slots[2 * pos + side] = value
                    nxt.append(2 * pos + side)
        level = nxt
    spelled = []

    def walk(pos, prefix):
        prefix = prefix + str(slots[pos])
        has_left = (2 * pos + 1) in slots
        has_right = (2 * pos + 2) in slots
        if not has_left and not has_right:
            spelled.append(prefix)
            return
        if has_left:
            walk(2 * pos + 1, prefix)
        if has_right:
            walk(2 * pos + 2, prefix)

    walk(0, "")
    return sum(int(text) for text in spelled)

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
4 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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