0968EasyTreeDepth-First SearchBreadth-First SearchBinary Tree

A Branch Whose Loads Add Up

Tracked in this browser only

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 rig hangs from a single top joint, and every joint carries at most two joints below it, held in a first slot and a second slot; either slot may be empty. Each joint carries a load, which may be negative.

The rig arrives as the flat list root, written level by level. root[0] is the top joint's load. Reading the list from left to right, every entry that is not null claims the next two unused positions as its first and second slot in that order, and null marks an empty slot. An entry that is null claims no positions of its own. If the list ends early, the slots that were never written are empty. An empty list means the rig has no joints at all.

A branch runs from the top joint down to a joint with both slots empty. Return true when some branch's loads add up to exactly targetSum.

Examples

Example 1

Input: root = [7, 4, 9, 2, null, null, 5], targetSum = 13
Output: true

The branch running 7, 4, 2 adds up to 13. The other branch runs 7, 9, 5, which comes to 21.

Example 2

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

The only branch runs 1 then 2, adding to 3. Note the top joint is not itself the end of a branch, since one of its slots is filled.

Example 3

Input: root = [], targetSum = 5
Output: false

The rig has no joints, so it has no branches and nothing can add up.

Constraints

The rig holds between 0 and 5000 joints.
0 <= root.length <= 15000
-1000 <= root[i] <= 1000
-1000 <= targetSum <= 1000
root[0] is not null when the list is not empty.

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 has_path_sum(root: list, targetSum: int) -> bool:

Java

public boolean hasPathSum(Integer[] root, int targetSum)

The intended approach

Time O(n) for n positions in the list: one pass to lay out the slots and one walk down.Space O(n) for the slots and the walk.

Turn the flat list into slots first: walk it level by level, where the filled joints of a level claim the next two positions each, in order, and record for every joint which positions hold its two slots. That is one pass over the list. Then walk down from the top joint carrying the loads added up so far. At a joint with both slots empty the branch has ended, so compare the running total against the target. Otherwise carry on into whichever slots are filled. Note that a joint with one slot filled is not the end of a branch, so its running total must not be compared. An empty rig has no branches at all, so nothing can add up.

Pythonentry point: has_path_sum

def has_path_sum(root, targetSum):
    if not root:
        return False
    n = len(root)
    left = [-1] * n
    right = [-1] * n
    level = [0]
    at = 1
    while level and at < n:
        nxt = []
        for owner in level:
            if at < n:
                if root[at] is not None:
                    left[owner] = at
                    nxt.append(at)
                at += 1
            if at < n:
                if root[at] is not None:
                    right[owner] = at
                    nxt.append(at)
                at += 1
        level = nxt
    found = [False]

    def walk(node, running):
        if found[0]:
            return
        running += root[node]
        if left[node] < 0 and right[node] < 0:
            if running == targetSum:
                found[0] = True
            return
        if left[node] >= 0:
            walk(left[node], running)
        if right[node] >= 0:
            walk(right[node], running)

    walk(0, 0)
    return found[0]

The obvious version

Lay the slots out the same way, then collect every branch as its own list of loads, adding each one up afterwards and looking for the target. Builds all the branches in full rather than carrying a running total.

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_has_path_sum(root, targetSum):
    if not root:
        return False
    n = len(root)
    kids = {}
    level = [0]
    at = 1
    while level and at < n:
        nxt = []
        for owner in level:
            kids.setdefault(owner, [])
            for _ in range(2):
                if at >= n:
                    break
                if root[at] is not None:
                    kids[owner].append(at)
                    nxt.append(at)
                at += 1
        level = nxt
    branches = []

    def gather(node, sofar):
        sofar = sofar + [root[node]]
        below = kids.get(node, [])
        if not below:
            branches.append(sofar)
            return
        for child in below:
            gather(child, sofar)

    gather(0, [])
    for branch in branches:
        total = 0
        for v in branch:
            total += v
        if total == targetSum:
            return True
    return False

How this problem was checked

Reference solution executed against 13 test cases — 13 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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in