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72. Path Sum II

mediumAsked at Vercel

Return all root-to-leaf paths where the sum equals target. Vercel asks this for the backtracking-with-running-sum pattern — same shape as their cost-bounded route discovery in the deployment graph.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Vercel loops.

  • Glassdoor (2025-12)— Vercel platform onsite; backtracking expected.
  • Blind (2026-Q1)— Listed in Vercel screen pool.

Problem

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references. A root-to-leaf path is a path starting from the root and ending at any leaf node.

Constraints

  • The number of nodes in the tree is in the range [0, 5000].
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000

Examples

Example 1

Input
root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output
[[5,4,11,2],[5,8,4,5]]

Example 2

Input
root = [1,2,3], targetSum = 5
Output
[]

Approaches

1. DFS collecting all root-to-leaf paths, filter

Collect every path; keep those with sum == target.

Time
O(n * h)
Space
O(n)
// O(n) paths in the worst case, each O(h). Better to filter on the fly.

Tradeoff: Wastes work emitting non-matching paths.

2. Backtracking with running remaining (optimal)

Track remaining = target - sum-so-far. At leaves, push path if remaining == 0. Use mutable path with push/pop for O(h) extra storage per path.

Time
O(n^2) worst-case output
Space
O(h)
function pathSum(root, targetSum) {
  const out = [];
  const path = [];
  function dfs(node, remaining) {
    if (!node) return;
    path.push(node.val);
    if (!node.left && !node.right && remaining === node.val) {
      out.push([...path]);
    }
    dfs(node.left, remaining - node.val);
    dfs(node.right, remaining - node.val);
    path.pop();
  }
  dfs(root, targetSum);
  return out;
}

Tradeoff: The mutable path with push/pop is the canonical backtracking pattern. Copying via spread on emit is the only allocation beyond the result itself.

Vercel-specific tips

Vercel grades the backtracking with mutable path. Bonus signal: emphasizing the leaf check (BOTH children null) rather than 'no left subtree' alone. Also flag that values can be negative — early-exit pruning based on remaining doesn't work.

Common mistakes

  • Pushing path WITHOUT spread on emit — collects a reference that mutates after backtrack.
  • Defining 'leaf' as 'no left child' — wrong, must be both null.
  • Trying to prune on remaining < 0 — fails when negative values exist.

Follow-up questions

An interviewer at Vercel may pivot to one of these next:

  • Path Sum (LC 112) — only boolean answer.
  • Path Sum III (LC 437) — paths anywhere in the tree.
  • Binary Tree Paths (LC 257) — all root-to-leaf paths without the sum constraint.

Solve it now

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Output

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FAQ

Why push then pop?

The path array is shared across all recursive calls. Push before recursion, pop after — this restores the array for the sibling subtree, exactly the backtracking idiom.

Why spread on emit?

If you push the same array reference, the post-emit pops would corrupt the saved path. Spreading captures a snapshot.

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