How do you solve Binary Tree Maximum Path Sum on LeetCode?

Binary Tree Maximum Path Sum (LeetCode #124) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: The maximum path sum can be found by considering paths that may not pass through the root.

What is the brute force solution for Binary Tree Maximum Path Sum?

The brute force approach for Binary Tree Maximum Path Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the path sum for every possible path in the binary tree. This means exploring all nodes and their possible paths, which can be quite inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Binary Tree Maximum Path Sum?

Binary Tree Maximum Path Sum uses the following concepts: Dynamic Programming, Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Dynamic Programming.

What are the interview tips for Binary Tree Maximum Path Sum?

Always clarify whether the path must pass through the root or not. Think about edge cases, such as trees with only one node or all negative values. Practice writing recursive functions, as they are commonly used in tree problems.

What are common mistakes when solving Binary Tree Maximum Path Sum?

Not considering paths that do not pass through the root. Failing to update the maximum path sum correctly during the traversal.

#124

Binary Tree Maximum Path Sum

Hard

Dynamic ProgrammingTreeDepth-First SearchBinary TreeDepth-First SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

💡

Intuition

Time O(n)Space O(h)

The optimal solution uses depth-first search (DFS) to calculate the maximum path sum efficiently. It keeps track of the maximum path sum at each node while ensuring that we only consider paths that contribute positively to the sum.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to keep track of the maximum path sum found.
2Step 2: Use DFS to traverse the tree, calculating the maximum contribution of each node to the path sum.
3Step 3: For each node, calculate the maximum path sum that can be formed by including its left and right children.
4Step 4: Update the global maximum path sum if the current node's path sum exceeds it.

solution.py22 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def maxPathSum(self, root: TreeNode) -> int:
10        max_sum = float('-inf')
11
12        def dfs(node):
13            nonlocal max_sum
14            if not node:
15                return 0
16            left = max(dfs(node.left), 0)
17            right = max(dfs(node.right), 0)
18            max_sum = max(max_sum, left + right + node.val)
19            return node.val + max(left, right)
20
21        dfs(root)
22        return max_sum

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(h) due to the recursive call stack, where h is the height of the tree.

1The maximum path sum can be found by considering paths that may not pass through the root.
2Using DFS allows us to explore all potential paths efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.