Bird
Raised Fist0
DSA Pythonprogramming~5 mins

DP on Trees Maximum Path Sum in DSA Python - Time & Space Complexity

Choose your learning style10 modes available
Time Complexity: DP on Trees Maximum Path Sum
O(n)
Understanding Time Complexity

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

How does the number of steps change when the tree has more nodes?

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 dfs(node):
        nonlocal max_sum
        if not node:
            return 0
        left = max(dfs(node.left), 0)
        right = max(dfs(node.right), 0)
        max_sum = max(max_sum, node.val + left + right)
        return node.val + max(left, right)
    dfs(root)
    return max_sum
    

This code finds the maximum sum of values along any path in a binary tree using a depth-first search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Recursive depth-first search (dfs) visits each node once.
  • How many times: Each node in the tree is visited exactly one time.
How Execution Grows With Input

As the number of nodes in the tree increases, the number of steps grows roughly the same way.

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

Pattern observation: The work grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum path sum grows directly in proportion to 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 work happens.

Interview Connect

Understanding this helps you solve many tree problems efficiently by using recursion smartly without repeating work.

Self-Check

"What if the tree was very unbalanced, 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?
Maximum path sum means:
easy
A. The largest sum of node values along any path between two nodes
B. The sum of all node values in the tree
C. The maximum value of a single node in the tree
D. The longest path length in the tree

Solution

  1. Step 1: Understand the path definition

    A path in a tree is any sequence of nodes connected by edges, not necessarily starting at the root.
  2. Step 2: Define maximum path sum meaning

    The maximum path sum is the highest sum of node values along any such path, which can start and end anywhere.
  3. Final Answer:

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

    Maximum path sum = largest node value sum [OK]
Hint: Maximum path sum is about sums, not counts or single nodes [OK]
Common Mistakes:
  • Confusing path sum with total tree sum
  • Thinking path must start at root
  • Considering only single nodes
2. Which of the following Python function signatures is correct for a recursive helper to find maximum path sum in a binary tree node?
def max_path_sum_helper(node: Optional[TreeNode]) -> int:
easy
A. def max_path_sum_helper(node: TreeNode) -> int:
B. def max_path_sum_helper(node: int) -> TreeNode:
C. def max_path_sum_helper(node: Optional[TreeNode]) -> int:
D. def max_path_sum_helper(node: TreeNode) -> None:

Solution

  1. Step 1: Understand parameter type

    The node can be None (empty), so Optional[TreeNode] is correct to allow None.
  2. Step 2: Understand return type

    The helper returns an integer representing max path sum from that node, so return type is int.
  3. Final Answer:

    def max_path_sum_helper(node: Optional[TreeNode]) -> int: -> Option C
  4. Quick Check:

    Optional node input and int output [OK]
Hint: Use Optional for nullable nodes and int return for sums [OK]
Common Mistakes:
  • Not allowing None as input node
  • Returning wrong type like None or TreeNode
  • Using wrong parameter types
3. What is the output 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 max_path_sum(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(max_path_sum(root))
medium
A. 35
B. 42
C. 27
D. 20

Solution

  1. Step 1: Calculate max path sums for subtrees

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

    Root value is -10, so max path sum including root is max(9, 0) + -10 + max(35, 0) = 34, but max_sum is updated to 42 from right subtree.
  3. Final Answer:

    42 -> Option B
  4. Quick Check:

    Max path sum = 15+20+7 = 42 [OK]
Hint: Ignore negative sums by max with 0 to maximize path [OK]
Common Mistakes:
  • Adding negative paths instead of ignoring
  • Confusing return value with max_sum
  • Missing nonlocal max_sum update
4. Find the bug in this code for maximum path sum in a binary tree:
def max_path_sum(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 is not declared as nonlocal inside helper
B. helper returns wrong value for leaf nodes
C. Should return max_sum inside helper function
D. Missing base case for empty tree

Solution

  1. Step 1: Check variable scope of max_sum

    max_sum is assigned inside helper but declared outside, so it needs 'nonlocal' to modify outer variable.
  2. Step 2: Understand effect of missing nonlocal

    Without 'nonlocal', max_sum inside helper is local, so outer max_sum never updates, causing wrong result.
  3. Final Answer:

    max_sum is not declared as nonlocal inside helper -> Option A
  4. Quick Check:

    Use nonlocal to update outer max_sum [OK]
Hint: Declare nonlocal for outer variables modified inside nested functions [OK]
Common Mistakes:
  • Forgetting nonlocal keyword
  • Returning max_sum inside helper incorrectly
  • Ignoring base case for None nodes
5. Given a binary tree where nodes can have negative values, how would you modify the maximum path sum algorithm to handle the case where all node values are negative?
Choose the best approach:
hard
A. Add absolute values of nodes to maximize sum
B. Ignore negative values and return 0 as max path sum
C. Only consider paths starting at the root
D. Return the maximum single node value if all sums are negative

Solution

  1. Step 1: Understand negative values impact

    If all nodes are negative, the maximum path sum is the largest (least negative) single node value, since no path sums to 0 and longer paths sum to less.
  2. Step 2: Modify algorithm to track max single node

    Keep track of the maximum node value separately and return it if all path sums are negative.
  3. Final Answer:

    Return the maximum single node value if all sums are negative -> Option D
  4. Quick Check:

    Handle all-negative by max single node value [OK]
Hint: Track max node value separately to handle all negatives [OK]
Common Mistakes:
  • Returning 0 when all nodes are negative
  • Adding absolute values incorrectly
  • Restricting paths to root only
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard