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Maximum Path Sum in Binary Tree in DSA Python - Time & Space Complexity

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Time Complexity: Maximum Path Sum in Binary Tree
O(n)
Understanding Time Complexity

We want to know how the time needed to find the maximum path sum in a binary tree grows as the tree gets bigger.

Specifically, how does the number of steps change when the number of nodes increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def maxPathSum(root):
    max_sum = float('-inf')
    def helper(node):
        nonlocal max_sum
        if not node:
            return 0
        left = max(helper(node.left), 0)
        right = max(helper(node.right), 0)
        max_sum = max(max_sum, node.val + left + right)
        return node.val + max(left, right)
    helper(root)
    return max_sum
    

This code finds the maximum sum of values along any path in a binary tree, where a path can start and end at any nodes.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Recursive calls visiting each node once.
  • How many times: Each node is visited exactly one time during the recursion.
How Execution Grows With Input

As the number of nodes grows, the function visits each node once, so the total steps grow roughly in direct proportion to the number of nodes.

Input Size (n)Approx. Operations
10About 10 visits
100About 100 visits
1000About 1000 visits

Pattern observation: The work grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum path sum grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The recursion visits nodes multiple times, so the time is more than linear."

[OK] Correct: Each node is visited once, and results are combined on the way back, so no repeated full visits happen.

Interview Connect

Understanding this time complexity helps you explain how efficient your solution is and shows you can analyze recursive tree problems clearly.

Self-Check

"What if the tree was very unbalanced and looked like a linked list? How would the time complexity change?"

Practice

(1/5)
1. What does the "maximum path sum" in a binary tree represent?
easy
A. The sum of values from root to the deepest leaf
B. The largest sum of values along any path between two nodes
C. The maximum value of a single node in the tree
D. The sum of all node values in the tree

Solution

  1. Step 1: Understand the path definition

    A path can start and end at any nodes, not necessarily root or leaf.
  2. Step 2: Define maximum path sum

    It is the highest sum of node values along any such path in the tree.
  3. Final Answer:

    The largest sum of values along any path between two nodes -> Option B
  4. Quick Check:

    Maximum path sum = largest sum on any path [OK]
Hint: Remember path can start/end anywhere, not just root or leaf [OK]
Common Mistakes:
  • Thinking path must start at root
  • Confusing max node value with max path sum
  • Summing all nodes instead of a path
2. Which of the following Python function signatures is correct for a recursive helper to find maximum path sum in a binary tree node root?
easy
A. def maxPathSum(root: TreeNode) -> int:
B. def maxPathSum(root: int) -> TreeNode:
C. def maxPathSum(root: TreeNode) -> TreeNode:
D. def maxPathSum(root: int) -> int:

Solution

  1. Step 1: Identify input type

    The function takes a TreeNode object representing the root of the tree.
  2. Step 2: Identify return type

    The function returns an integer representing the maximum path sum.
  3. Final Answer:

    def maxPathSum(root: TreeNode) -> int: -> Option A
  4. Quick Check:

    Input=TreeNode, Output=int [OK]
Hint: Input is tree node, output is integer sum [OK]
Common Mistakes:
  • Using int as input instead of TreeNode
  • Returning TreeNode instead of int
  • Mixing input and output types
3. Given the following binary tree and code snippet, what is the output?
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def maxPathSum(root):
    max_sum = float('-inf')
    def helper(node):
        nonlocal max_sum
        if not node:
            return 0
        left = max(helper(node.left), 0)
        right = max(helper(node.right), 0)
        max_sum = max(max_sum, node.val + left + right)
        return node.val + max(left, right)
    helper(root)
    return max_sum

root = TreeNode(-10, TreeNode(9), TreeNode(20, TreeNode(15), TreeNode(7)))
print(maxPathSum(root))
medium
A. 35
B. 27
C. 42
D. 20

Solution

  1. Step 1: Trace left and right subtree sums

    Left subtree max path sum is 9, right subtree max path sum is 20 + 15 + 7 = 42.
  2. Step 2: Calculate max path sum including root

    Root (-10) + left (9) + right (20 + 15 + 7) = -10 + 9 + 22 = 21, but max path sum is updated at right subtree as 42.
  3. Step 3: Identify maximum path sum

    The maximum path sum is 42, but the path that includes root is less than 42.
  4. Final Answer:

    42 -> Option C
  5. Quick Check:

    Max path sum = 42 [OK]
Hint: Add max(0, left/right) to avoid negative sums [OK]
Common Mistakes:
  • Including negative sums from subtrees
  • Confusing return value with max_sum
  • Not using nonlocal for max_sum
4. Find the error in the following code snippet for maximum path sum in a binary tree:
def maxPathSum(root):
    max_sum = float('-inf')
    def helper(node):
        if not node:
            return 0
        left = helper(node.left)
        right = helper(node.right)
        max_sum = max(max_sum, node.val + left + right)
        return node.val + max(left, right)
    helper(root)
    return max_sum
medium
A. max_sum should be declared as nonlocal inside helper
B. Return statement should be outside helper function
C. Base case should return None instead of 0
D. max_sum should be initialized to 0 instead of -inf

Solution

  1. Step 1: Identify variable scope issue

    max_sum is assigned inside helper but declared outside, so Python treats it as local inside helper.
  2. Step 2: Fix scope with nonlocal keyword

    Adding 'nonlocal max_sum' inside helper allows updating the outer max_sum variable.
  3. Final Answer:

    max_sum should be declared as nonlocal inside helper -> Option A
  4. Quick Check:

    Use nonlocal to update outer variable [OK]
Hint: Use nonlocal to modify outer variables inside nested functions [OK]
Common Mistakes:
  • Forgetting nonlocal causes UnboundLocalError
  • Changing base case return value incorrectly
  • Initializing max_sum to 0 misses negative sums
5. You want to find the maximum path sum in a binary tree where some nodes have negative values. Which approach correctly handles negative values to find the maximum path sum?
hard
A. Only consider paths starting at root to leaf
B. Add all child sums regardless of sign to current node value
C. Replace negative node values with zero before calculation
D. Ignore negative child sums by taking max(child_sum, 0) before adding to current node

Solution

  1. Step 1: Understand impact of negative values

    Negative sums reduce total path sum, so ignoring them improves max sum.
  2. Step 2: Use max(child_sum, 0) to ignore negatives

    This ensures only positive contributions add to current node's path sum.
  3. Step 3: Avoid altering original node values

    Replacing negatives with zero changes tree data and is incorrect.
  4. Final Answer:

    Ignore negative child sums by taking max(child_sum, 0) before adding to current node -> Option D
  5. Quick Check:

    Ignore negatives with max(child_sum, 0) [OK]
Hint: Use max(child_sum, 0) to skip negative paths [OK]
Common Mistakes:
  • Adding negative sums reduces max path sum
  • Modifying node values instead of sums
  • Restricting paths to root-leaf only
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