DP on Trees Maximum Path Sum in DSA C++ - Execution Trace

Concept Flow - DP on Trees Maximum Path Sum

Start at root node

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Compute max path sum from left subtree

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Compute max path sum from right subtree

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Calculate max path through current node

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Update global max if current path is higher

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Return max path sum including current node and one subtree

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Repeat for all nodes via recursion

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Final global max is maximum path sum in tree

We recursively compute max path sums from left and right subtrees, update a global max including paths through current node, and return max sums for parent computations.

Execution Sample

DSA C++

int maxPathSum(TreeNode* root) {
  int globalMax = INT_MIN;
  maxGain(root, globalMax);
  return globalMax;
}

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

This code finds the maximum path sum in a binary tree by recursively calculating max gains from subtrees and updating a global maximum.

Execution Table

Step	Operation	Node Visited	Left Max Gain	Right Max Gain	Global Max Updated	Return Value	Visual State
1	Visit leaf node	Node 4 (val=4)	0	0	4	4	4
2	Visit leaf node	Node 5 (val=5)	0	0	5	5	5
3	Visit node 2	Node 2 (val=2)	4	5	11	7	2 -> 4, 5
4	Visit leaf node	Node 3 (val=3)	0	0	11	3	3
5	Visit root node	Node 1 (val=1)	7	3	11	8	1 -> 2 -> 4,5 and 3
6	End	None	-	-	11	-	Final max path sum = 11

💡 All nodes visited, recursion unwound, global max holds maximum path sum

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	Final
globalMax	-∞	4	5	11	11	11	11
Return Value (maxGain)	-	4	5	7	3	8	-

Key Moments - 3 Insights

Why do we take max(0, maxGain(...)) for left and right subtrees?

Because negative path sums reduce the total, we ignore them by taking max with 0. See execution_table rows 1 and 2 where leaf nodes return positive gains, ensuring only positive contributions add to global max.

Why update globalMax with node->val + left + right instead of just one side?

Because the max path can pass through the current node connecting left and right subtrees. This is shown in step 3 where globalMax updates to 11 by summing node 2's value and both children gains.

What does the return value represent in recursion?

It represents the max gain from the current node to one subtree path (left or right), used by parent nodes to build paths. See step 5 where return value 8 is node 1's val plus max(left=7, right=3).

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the globalMax value after visiting node 2?

A 9

B 7

C 11

D 5

💡 Hint

Check the 'Global Max Updated' column in execution_table row 3

At which step does the recursion return the value 8?

A Step 5

B Step 3

C Step 4

D Step 2

💡 Hint

Look at the 'Return Value' column in execution_table row 5

If node 5 had value -1 instead of 5, how would the globalMax after step 3 change?

A 11

B 6

C It would increase

D It would become 5

💡 Hint

Negative values reduce max gains, see how max(0, maxGain) affects globalMax in variable_tracker

Concept Snapshot

DP on Trees Maximum Path Sum:
- Use recursion to get max gain from left and right children
- Ignore negative gains by max(0, gain)
- Update global max with node value + left + right gains
- Return node value + max(left, right) for parent use
- Final global max is max path sum anywhere in tree

Full Transcript

This concept uses dynamic programming on trees to find the maximum path sum. We recursively visit each node, calculate the maximum gain from left and right subtrees ignoring negative sums, update a global maximum path sum that may pass through the current node, and return the maximum gain including the current node to its parent. The global maximum after visiting all nodes is the answer. Key points include ignoring negative paths, updating global max with both children plus current node, and returning max gain for parent recursion.

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