Is Binary Tree Maximum Path Sum easy or hard?

Binary Tree Maximum Path Sum is classified as a Hard problem because it requires understanding two different path concepts: the extendable path returned to the parent and the complete path passing through a node. Handling negative values and combining subtree results correctly makes it challenging.

Binary Tree Maximum Path Sum Python/Java solution

Python and Java implementations both use recursive DFS. The helper function returns the maximum gain from the current node while updating a global maximum path sum. Both implementations run in O(n) time and require O(h) stack space.

How to solve Binary Tree Maximum Path Sum in O(n)?

Perform a postorder DFS traversal and compute the maximum gain from each node. For each node, calculate left_gain and right_gain, ignoring negative contributions by clamping them to zero. Update a global maximum with node.val + left_gain + right_gain, and return node.val + max(left_gain, right_gain) to the parent.

What is the best approach for Binary Tree Maximum Path Sum?

The optimal approach uses depth-first search with dynamic programming on the tree. During a postorder traversal, compute the maximum gain from each child and update a global maximum path sum at every node. This solution processes each node once, giving O(n) time complexity and O(h) recursion stack space.

Is Binary Tree Maximum Path Sum asked at Google/Amazon/Meta?

Binary Tree Maximum Path Sum is a well-known hard tree problem frequently reported in interviews at companies like Google, Amazon, Meta, and Microsoft. It tests depth-first search, recursion, and the ability to reason about paths in trees with negative values.

What data structure is used in Binary Tree Maximum Path Sum?

The problem operates on a binary tree data structure. The solution typically uses recursion with depth-first search and maintains a global variable to track the maximum path sum while traversing the tree.

What is the time complexity of Binary Tree Maximum Path Sum?

The time complexity is O(n) where n is the number of nodes in the binary tree. Each node is visited exactly once during the DFS traversal. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

Description Solution Video 15:19

124. Binary Tree Maximum Path Sum

Hard42.7% acceptance25 companies

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node's values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

Example 1:

Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.

Example 2:

Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.

Constraints:

The number of nodes in the tree is in the range [1, 3 * 10⁴].
-1000 <= Node.val <= 1000

Asked at top 20 of 25 companies

Baidu Adobe DoorDash Wix Ola Amazon Tiktok Salesforce Citadel Nutanix Datadog

Google

+8 more

Microsoft Meta Oracle Yandex Snapchat Bloomberg Uber Apple

Topics

15:19

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

124. Binary Tree Maximum Path Sum

Asked at top 20 of 25 companies

Topics

Similar problems