What is the brute force approach and complexity for Binary Tree Maximum Path Sum?

Treat each node as the top of the path, and search each of its sides from scratch with time complexity O(N²) and space complexity O(H).

What is the optimal approach and complexity for Binary Tree Maximum Path Sum?

One post-order pass: each node returns its best one-side gain and records the best path that bends there with optimal time complexity O(N) and space complexity O(H).

How do you identify the pattern for Binary Tree Maximum Path Sum?

Look for key signals in the prompt: "path can start and end at any node" indicates Not root-to-leaf. The best path can bend at any node, so every node is a possible top.; "node values can be negative" indicates A side with a negative sum must be dropped. This is the max(0, ·) clamp.; "largest path sum" indicates Maximise a sum over paths. Each node returns a value upward and records a candidate on the side..

Binary Tree Maximum Path Sum

Hard·LeetCode #124 Pro OnlyInterview-Level

LeetCode

Problem Statement

You are given the root of a binary tree. A path is a sequence of nodes where each pair of neighbours is joined by an edge. Each node can appear in a path at most once.

A path does not have to start or end at the root, and it does not have to go through the root. Its path sum is the total of the node values on it. Node values can be negative.

Return the largest path sum over all paths. A path must contain at least one node.

Input

The root of a binary tree. It has between 1 and 3 × 10^4 nodes. Values can be negative.

Output

The largest path sum of any non-empty path.

Example 1

Input: root = [1, 2, 3]

Output: 6

The best path is 2 → 1 → 3, with sum 2 + 1 + 3 = 6. It goes through the root.

Example 2

Input: root = [-10, 9, 20, null, null, 15, 7]

Output: 42

The best path is 15 → 20 → 7, with sum 42. It does not include the root, which has value −10.

Example 3

Input: root = [1, 20, null, 15, 7]

Output: 42

The best path is still 15 → 20 → 7. It cannot also continue up to 1, because a path that goes through 20 on both sides cannot go up to 20's parent.

Constraints

• The number of nodes is in the range [1, 3 × 10^4].
• −1000 ≤ Node.val ≤ 1000

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“path can start and end at any node”

→ Not root-to-leaf. The best path can bend at any node, so every node is a possible top.

“node values can be negative”

→ A side with a negative sum must be dropped. This is the max(0, ·) clamp.

“largest path sum”

→ Maximise a sum over paths. Each node returns a value upward and records a candidate on the side.

Approach Complexity Ladder

Brute Force — best downward gains at every node, each with a separate searchO(N²)O(H)

Treat each node as the top of the path, and search each of its sides from scratch

Optimal — return one side upward, check both sides at each nodeO(N)O(H)

One post-order pass: each node returns its best one-side gain and records the best path that bends there

Brute Force Intuition

Every path has a top node where it turns. For each node, find the best downward sum on its left side and on its right side, using a separate search for each side. Add the node's value and both sides (a side with a negative sum is dropped). The largest total over all nodes is the answer.

Recognition rule for this sub-pattern:

Each node returns a summary of its subtree (height, best path). The global answer is updated while combining.

Remember this · Tree DFS

Trigger

Paths, heights, diameters, validating structure, ancestors, anything "for every subtree".

Think

If my left and right children already gave me their answers, how do I build mine?

Key question

What does my recursive function return, and what do I pass down?

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