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

Time Complexity: Maximum Path Sum in Binary Tree

O(n)

Understanding Time Complexity

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

The question is: 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.

int maxPathSumHelper(TreeNode* node, int& maxSum) {
    if (!node) return 0;
    int left = max(0, maxPathSumHelper(node->left, maxSum));
    int right = max(0, maxPathSumHelper(node->right, maxSum));
    maxSum = max(maxSum, left + right + node->val);
    return max(left, right) + node->val;
}

int maxPathSum(TreeNode* root) {
    int maxSum = INT_MIN;
    maxPathSumHelper(root, maxSum);
    return maxSum;
}
    

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.

How Execution Grows With Input

As the number of nodes in the tree increases, the function visits each node once, so the total steps grow directly with the number of nodes.

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

Pattern observation: The steps increase in a straight line as the tree grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum path sum grows linearly 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 work happens.

Interview Connect

Understanding this linear time complexity helps you explain efficient tree traversal and recursion skills clearly in interviews.

Self-Check

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

Practice

1. What does the term Maximum Path Sum in a binary tree mean?

easy

A. The sum of all node values in the tree

B. The largest sum of values along any path between two nodes in the tree

C. The maximum value of a single node in the tree

D. The sum of values from root to the deepest leaf only

Solution

1. Step 1: Understand the path definition

   A path in a binary tree 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 in the tree -> Option B

4. Quick Check:

   Maximum path sum means highest sum on any path [OK]

Hint: Maximum path sum is any path, not just root to leaf [OK]

Common Mistakes:

• Confusing maximum path sum with sum of all nodes
• Thinking path must start at root
• Considering only leaf nodes for path ends

2. Which of the following C++ function signatures is correct for a helper function that returns the maximum path sum starting from a given node in a binary tree?

easy

A. int maxPathSumHelper(TreeNode* root, int& maxSum);

B. void maxPathSumHelper(TreeNode root, int maxSum);

C. int maxPathSumHelper(TreeNode root);

D. int maxPathSumHelper(TreeNode* root);

Solution

1. Step 1: Understand parameter passing

   The function needs to update maxSum globally, so it must be passed by reference.

2. Step 2: Check pointer usage

   TreeNode* is used to pass node pointers, not objects directly.

3. Final Answer:

   int maxPathSumHelper(TreeNode* root, int& maxSum); -> Option A

4. Quick Check:

   Use pointer and reference for maxSum [OK]

Hint: Pass maxSum by reference and node as pointer [OK]

Common Mistakes:

• Passing TreeNode by value instead of pointer
• Not passing maxSum by reference
• Using void return type instead of int

3. Consider the following binary tree:

    1
   / \
  2   3

What is the maximum path sum returned by the standard algorithm?

medium

A. 3

B. 5

C. 4

D. 6

Solution

1. Step 1: Calculate path sums including root

   Paths: 2->1->3 sum = 6, 2->1 sum = 3, 1->3 sum = 4.

2. Step 2: Identify maximum sum

   The maximum path sum is 6 from path 2->1->3.

3. Final Answer:

   6 -> Option D

4. Quick Check:

   2+1+3 = 6 [OK]

Hint: Check all paths including both children and root [OK]

Common Mistakes:

• Considering only root-to-leaf paths
• Ignoring paths passing through root
• Adding only one child value

4. Given this buggy code snippet for maximum path sum, what is the main error?

int maxPathSumHelper(TreeNode* root, int& maxSum) {
  if (!root) return 0;
  int left = maxPathSumHelper(root->left, maxSum);
  int right = maxPathSumHelper(root->right, maxSum);
  int current = root->val + left + right;
  maxSum = max(maxSum, current);
  return root->val + max(left, right);
}

What is the bug?

medium

A. It does not ignore negative path sums

B. It returns sum of both children instead of max child

C. It updates maxSum incorrectly

D. It does not handle null root

Solution

1. Step 1: Analyze negative path sums handling

   The code adds left and right directly without ignoring negative sums, which can reduce maxSum.

2. Step 2: Correct approach for negatives

   Should use max(0, left) and max(0, right) to ignore negative sums.

3. Final Answer:

   It does not ignore negative path sums -> Option A

4. Quick Check:

   Negative sums must be ignored to maximize path sum [OK]

Hint: Use max(0, child) to ignore negative sums [OK]

Common Mistakes:

• Adding negative child sums reduces maxSum
• Returning sum of both children instead of max child
• Not checking for null root

5. Given the binary tree below, what is the maximum path sum?

       -10
       /  \
      9   20
          / \
         15  7

Choose the correct maximum path sum.

hard

A. 27

B. 35

C. 42

D. 15

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: Consider paths through root

   Path through root: -10 + max(0,9) + max(0,15+20+7) = -10 + 9 + 42 = 41, less than 42.

3. Step 3: Identify maximum path sum

   The maximum path sum is 42 from path 15->20->7.

4. Final Answer:

   42 -> Option C

5. Quick Check:

   15+20+7 = 42 [OK]

Hint: Check all subtree sums and paths through root [OK]

Common Mistakes:

• Including negative root value in max path sum
• Ignoring paths not passing through root
• Adding left and right children without max(0, val)