Of course! This is a great idea. Documenting your code well is just as important as writing code that works, especially in a team setting. Your algorithm is a classic and very elegant solution to this problem.
Let's break it down. I'll provide a fully commented version of your code and then walk through the logic step-by-step, just like I would for a tutoring session.
Here is your code, enhanced with docstrings and inline comments. This is what you could submit for your project.
from typing import Optional
class TreeNode:
"""A standard definition for a binary tree node."""
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution: # It's good practice to put methods in a class
def maxPathSum(self, root: Optional[TreeNode]) -> int:
"""
Calculates the maximum path sum in a binary tree.
A path is defined as any sequence of nodes from some starting node to any
node in the tree along the parent-child connections. The path must contain
at least one node and does not need to pass through the root.
Args:
root: The root node of the binary tree.
Returns:
The maximum path sum found in the tree.
"""
# This variable will store the maximum path sum found *anywhere* in the
# tree. We initialize it to negative infinity to handle cases with
# negative node values correctly. It's a "global" variable for our recursion.
self.max_path_sum = float('-inf')
# We use a helper function (DFS) to traverse the tree.
self._dfs(root)
return self.max_path_sum
def _dfs(self, node: Optional[TreeNode]) -> int:
"""
A Depth-First Search helper function that does two things:
1. It calculates the maximum path sum that can be formed *through* the current `node`.
This path looks like an inverted 'V' (left_child -> node -> right_child).
It updates the overall `self.max_path_sum` if this path is the new maximum.
2. It returns the maximum path sum that *starts* at the current `node` and
goes downwards in a straight line (either through the left or right child).
This is the value its parent node will use.
Args:
node: The current node in the traversal.
Returns:
The maximum path sum starting at `node` and extending downwards.
"""
# Base case: A null node contributes 0 to a path sum.
if not node:
return 0
# Recursively find the max path sum starting from the left and right children.
# We use max(..., 0) because we don't want to include paths with a negative
# sum. If a child's path sum is negative, it's better to not include that
# child's path at all (i.e., treat its contribution as 0).
left_gain = max(self._dfs(node.left), 0)
right_gain = max(self._dfs(node.right), 0)
# --- This is where we check for the overall maximum path sum ---
# A path can "turn" at the current node, using its left and right subtrees.
# This path (left_gain + node.val + right_gain) is a candidate for the
# overall maximum path sum. It cannot be extended further up to the parent
# because a path can't split.
current_path_sum = node.val + left_gain + right_gain
self.max_path_sum = max(self.max_path_sum, current_path_sum)
# --- This is the value we return to the parent node ---
# For the recursion, a node can only pass up a path that goes in one
# direction (a straight line). So, we return the node's value plus the
# larger of the gains from its left or right child.
return node.val + max(left_gain, right_gain)
Self-correction: I changed ans to self.max_path_sum and the nested function to a private method _dfs. This is a more standard object-oriented way to handle this in Python, avoiding the nonlocal keyword which can sometimes be less clear.
Tutoring Session: Explaining the "Why"
Great, now let's talk through why it works. This is the part that will help you explain it to your professor and teammates.
The Core Problem
The trickiest part of this problem is that a "path" can start and end anywhere.
- It could be a straight line from the root down to a leaf.
- It could be from a leaf up to a node and then down to another leaf (forming an inverted 'V').
- It could be just a single node.
Our algorithm needs to check every single possibility. A brute-force approach would be way too slow.
The "Aha!" Moment: The Dual-Purpose Recursive Function
Your solution is so clever because the dfs (or _dfs in my refactored version) function does two different but related jobs at the same time.
Let's focus on what happens at any single node, say N.
Job 1: Find the best path that turns at N
Imagine a path that looks like this: ... -> Left Child -> N -> Right Child -> ...
This path has its highest point, or its "peak," at node N. The sum of this path would be (best path from left child) + N.val + (best path from right child).
This is exactly what this line does:
ans = max(ans, root.val + left + right)
This value is a candidate for the overall maximum path sum in the entire tree. We compare it with our global ans and update ans if we've found a new best. But this path cannot be extended upwards to N's parent, because that would mean visiting N twice. So, we just update our global ans and we're done with this potential path.
Job 2: Report the best straight path upwards
Now, N's parent needs to do its own calculations. When N's parent looks at N, it can only form a path that goes through N in a straight line, like Parent -> N -> N's Child.
So, N needs to report back to its parent: "The best possible path you can build that starts with me and goes downwards has a sum of X."
This X is N.val + max(best path from left child, best path from right child). It can't include both children, because that would be a 'V' shape, not a straight path.
This is exactly what this line does:
return root.val + max(left, right)
The Crucial max(..., 0)
Why do we do left = max(dfs(root.left), 0)?
Imagine the best path extending from a child node has a negative sum (e.g., a subtree of [-5, -2, -10]). If we are forced to include it, it will only decrease our total path sum. It's always better to just not include that path. By taking max(path_sum, 0), we are essentially saying, "If the best path from this child is negative, I'll just stop here and pretend the path from that child has a sum of 0."
A Quick Walkthrough
Let's use this tree:
-10
/ \
9 20
/ \
15 7
dfs(15): left=0, right=0.
ans update: max(ans, 15+0+0). ans becomes 15.- Returns:
15 + max(0,0) = 15.
dfs(7): left=0, right=0.
ans update: max(15, 7+0+0). ans is still 15.- Returns:
7 + max(0,0) = 7.
dfs(20):
- It receives
left_gain = 15 (from node 15) and right_gain = 7 (from node 7).
ans update (Job 1): max(ans, 20 + 15 + 7). max(15, 42). ans is now 42! This is our best path so far.- Returns (Job 2):
20 + max(15, 7) = 20 + 15 = 35. It tells its parent (-10) that the best path starting from 20 has a sum of 35.
dfs(9): left=0, right=0.
ans update: max(42, 9+0+0). ans is still 42.- Returns:
9 + max(0,0) = 9.
dfs(-10) (the root):
- It receives
left_gain = 9 (from node 9) and right_gain = 35 (from node 20).
ans update (Job 1): max(ans, -10 + 9 + 35). max(42, 34). ans is still 42.- Returns (Job 2):
-10 + max(9, 35) = -10 + 35 = 25. (This value isn't used by anyone, but it's what the function returns).
Finally, the initial call to maxPathSum finishes, and it returns the final ans, which is 42.
Hope this detailed breakdown helps! You've written a very solid piece of code. This documentation and explanation should make it crystal clear to everyone on your team. Good luck with the project
