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

Time Complexity: DP on Trees Maximum Path Sum

O(n)

Understanding Time Complexity

We want to understand how the time needed grows when finding the maximum path sum in a tree using dynamic programming.

How does the number of steps change as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

This code finds the maximum sum of values along any path in a binary tree, using recursion and dynamic programming.

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 tree grows, the number of nodes increases, and the function visits each node once.

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

Pattern observation: The work grows directly with the number of nodes, so doubling nodes doubles work.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows linearly with the number of nodes in the tree.

Common Mistake

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

[OK] Correct: Each node is visited once, and results are combined without repeated visits, so time stays linear.

Interview Connect

Understanding this linear time complexity helps you explain efficient tree algorithms clearly and confidently in interviews.

Self-Check

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

Practice

1. What does the "Maximum Path Sum" in a tree represent in the context of dynamic programming on trees?

easy

A. The largest sum of node values on any path between two nodes in the tree

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 problem definition

   The maximum path sum is the highest sum of values collected by traveling along any path between two nodes in the tree, not necessarily passing through the root.

2. Step 2: Differentiate from other sums

   It is not the sum of all nodes or just the maximum node value, but the maximum sum along any connected path.

3. Final Answer:

   The largest sum of node values on any path between two nodes in the tree -> Option A

4. Quick Check:

   Maximum path sum = largest sum on any node-to-node path [OK]

Hint: Maximum path sum is about any path, not whole tree sum [OK]

Common Mistakes:

• Confusing maximum path sum with total sum of all nodes
• Thinking it is the maximum single node value
• Assuming path must start at root

2. Which of the following C++ function signatures is correct for a recursive function that computes the maximum path sum in a binary tree node structure defined as struct TreeNode { int val; TreeNode* left; TreeNode* right; };?

easy

A. int maxPathSum(TreeNode root);

B. void maxPathSum(TreeNode root);

C. int maxPathSum(TreeNode* root);

D. int maxPathSum(TreeNode* root, int sum);

Solution

1. Step 1: Check parameter passing

   The tree nodes should be passed as pointers to avoid copying the entire subtree, so TreeNode* is correct.

2. Step 2: Check return type and parameters

   The function returns an integer representing the maximum path sum, so int return type is correct. Additional parameters like int sum are not necessary for the main function signature.

3. Final Answer:

   int maxPathSum(TreeNode* root); -> Option C

4. Quick Check:

   Use pointer parameter and int return type [OK]

Hint: Use pointer parameter and int return type for tree recursion [OK]

Common Mistakes:

• Passing TreeNode by value causing copies
• Using void return type when int is needed
• Adding unnecessary parameters in main signature

3. Consider the following C++ code snippet for maximum path sum in a binary tree:

struct TreeNode { int val; TreeNode* left; TreeNode* right; };

int maxSum = INT_MIN;

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

int maxPathSum(TreeNode* root) {
  dfs(root);
  return maxSum;
}

// Tree structure:
//       1
//      / \
//     2   3
//
// What is the output of maxPathSum(root)?

medium

A. 5

B. 6

C. 3

D. 1

Solution

1. Step 1: Calculate dfs values for each node

   For leaf nodes 2 and 3, dfs returns their values (2 and 3). For root 1, left=2, right=3, so maxSum = max(INT_MIN, 1+2+3) = 6.

2. Step 2: Return maxSum after dfs completes

   The function returns maxSum = 6, which is the maximum path sum (2 -> 1 -> 3).

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Max path sum = 2 + 1 + 3 = 6 [OK]

Hint: Add node value plus max left and right sums, ignore negatives [OK]

Common Mistakes:

• Ignoring max(0, dfs(...)) causing negative sums
• Returning node->val + left + right instead of max path
• Confusing maxSum update with return value

4. The following code attempts to compute the maximum path sum in a binary tree but has a bug:

int maxSum = INT_MIN;

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

What is the main issue causing incorrect results?

medium

A. It does not handle null nodes properly

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

C. It does not update maxSum correctly

D. It does not ignore negative sums from children

Solution

1. Step 1: Analyze child sums usage

   The code uses raw dfs values for left and right without ignoring negative sums, which can reduce the path sum.

2. Step 2: Identify fix for negative sums

   To fix, use max(0, dfs(...)) to ignore negative sums from children, ensuring path sum only increases or stays same.

3. Final Answer:

   It does not ignore negative sums from children -> Option D

4. Quick Check:

   Use max(0, dfs(child)) to avoid negative sums [OK]

Hint: Always ignore negative child sums with max(0, dfs(child)) [OK]

Common Mistakes:

• Not using max(0, ...) causing negative path sums
• Returning sum of both children instead of max for recursion
• Incorrect maxSum update logic

5. Given a binary tree where nodes can have negative values, how would you modify the standard maximum path sum algorithm to correctly handle the case where the maximum path sum is a single node with a negative value? Choose the best approach.

hard

A. Initialize maxSum with INT_MIN and use max(0, dfs(child)) in recursion

B. Initialize maxSum with 0 and ignore negative node values completely

C. Initialize maxSum with INT_MIN and allow dfs to return negative values without max(0, ...)

D. Initialize maxSum with 0 and always add node values regardless of sign

Solution

1. Step 1: Recall the standard algorithm

   The standard algorithm initializes maxSum to INT_MIN and uses left = max(0, dfs(left)), right = max(0, dfs(right)), updating maxSum = max(maxSum, node.val + left + right). This captures single negative nodes as node.val + 0 + 0.

2. Step 2: Why it handles negatives

   max(0, dfs(child)) prevents extending to negative subtrees but allows dfs to return node.val + 0 (negative if node negative). INT_MIN init captures negative maxima.

3. Step 3: Why A fails

   Without max(0, ...), negative child sums pollute parent updates (e.g., root=-2, left=-3: maxSum wrongly stays -3 instead of -2).

4. Final Answer:

   Initialize maxSum with INT_MIN and use max(0, dfs(child)) in recursion -> Option A

5. Quick Check:

   max(0, dfs(child)) + INT_MIN handles negative single nodes [OK]

Hint: Standard max(0, dfs(child)) + INT_MIN handles negative nodes correctly [OK]

Common Mistakes:

• Omitting max(0, ...) letting negative children reduce parent path sums
• Initializing maxSum with 0 missing negative maxima
• Ignoring all negative nodes