Binary Tree Maximum Path Sum Leetcode Problem 124 [Python Solution]

In this blog post, we're going to tackle a challenging LeetCode problem known as Binary Tree Maximum Path Sum This problem falls under the category of trees and is categorized as a hard-level challenge.

We'll provide a Python solution that not only solves the problem but also explains the thought process behind it.

Problem Overview

Question: 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.

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

Given the root of a binary tree, your task is to 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 * 10^4].
• Node values are in the range [-1000, 1000].

To understand this problem, you need to find the maximum path sum in a binary tree.

A path is defined as a sequence of nodes where each pair of adjacent nodes has an edge connecting them.

However, the path doesn't necessarily need to pass through the root of the tree.

Understanding the Problem Constraints

Before we delve into the solution, it's essential to grasp the constraints provided in the problem:

1. The tree can have a large number of nodes, up to 30,000.
2. Node values can be any integer between -1000 and 1000.

Binary Tree Maximum Path Sum LeetCode Problem Solution

Now, let's explore the Python solution for this problem.

We'll take a recursive approach to solve it efficiently.

1. Bruteforce Approach

To solve this problem, we can traverse the binary tree and, for each node, calculate the maximum path sum by considering all possible paths through that node.

This would require a recursive function that explores both the left and right subtrees for each node.

Let's break down the steps involved:

1. We start at the root node.
2. For each node, we calculate the maximum path sum that can pass through it without splitting.

To do this, we take the maximum of the path sums from its left and right subtrees and add the node's value.
3. Next, we calculate the maximum path sum that can pass through the node while splitting it.

This involves taking the sum of the node's value, the maximum path sum from the left subtree (without splitting), and the maximum path sum from the right subtree (without splitting).
4. We update our result with the maximum path sum obtained by splitting the node if it's greater than the current result.
5. Finally, we return the maximum path sum without splitting for this node, which will be used when calculating the path sum for its parent node.

Here's the Python code implementing this approach:

def maxPathSum(root):
    result = [root.val]  # Store the result as a list for modification

    # Return the max path sum without splitting
    def dfs(root):
        if not root:
            return 0

        leftMax = dfs(root.left)
        rightMax = dfs(root.right)

        # Ensure negative values are not included in the path
        leftMax = max(leftMax, 0)
        rightMax = max(rightMax, 0)

        # Calculate max path sum WITH splitting the node
        result[0] = max(result[0], root.val + leftMax + rightMax)

        return root.val + max(leftMax, rightMax)

    dfs(root)
    return result[0]

This solution is efficient as it explores each node only once, resulting in a time complexity of O(N), where N is the number of nodes in the tree.

The memory complexity is determined by the tree's height, which is typically O(log N) for balanced trees.

Time and Space Complexity

To summarize the time and space complexity of this solution:

• Time Complexity: O(N) – We visit each node in the tree exactly once.
• Space Complexity: O(H) – The space used by the recursive call stack, where H is the height of the tree.

Reasoning Behind Our Approach

The key to solving this problem efficiently lies in understanding that, for each node, we need to compute two values:

1. The maximum path sum that can pass through the node without splitting.
2. The maximum path sum that can pass through the node while splitting it (i.e., the path can continue on either the left or the right subtree).

By comparing these two values, we can find the maximum path sum for the entire tree.

This approach ensures that we consider all possible paths in the tree while avoiding unnecessary recomputation by caching results for subproblems.

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Conclusion

In this blog post, we tackled the Binary Tree Maximum Path Sum LeetCode problem.

We provided a Python solution that leverages a recursive approach to efficiently find the maximum path sum in a binary tree.

This solution considers all possible paths while avoiding redundant calculations, resulting in a time complexity of O(N).

Remember that understanding and practicing tree-related problems is essential for improving your problem-solving skills in computer science and data structures.

We encourage you to comment, ask questions, make suggestions, and share this content with others who might find it helpful.

You can further explore this problem and submit your solution on LeetCode.

If you have any questions or need additional explanations, please feel free to ask.

Happy coding!