DP on Trees Maximum Path Sum in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the maximum path sum starting from the current node.

DSA C++

int maxPathDown(TreeNode* node) {
    if (node == nullptr) return 0;
    int left = max(0, maxPathDown(node->left));
    int right = max(0, maxPathDown(node->right));
    return node->val + [1];
}

Drag options to blanks, or click blank then click option'

Amax(left, right)

Bleft + right

Cleft - right

Dmin(left, right)

Attempts:

3 left

💡 Hint

Common Mistakes

Adding both left and right sums instead of choosing the maximum.

Using min instead of max.

Not handling negative sums by max with 0.

We want the maximum path sum starting from the current node going down one side, so we add the node's value to the maximum of left or right child sums.

2fill in blank

medium

Complete the code to update the global maximum path sum including the current node and both children.

DSA C++

int maxSum = INT_MIN;

int maxPathDown(TreeNode* node) {
    if (node == nullptr) return 0;
    int left = max(0, maxPathDown(node->left));
    int right = max(0, maxPathDown(node->right));
    maxSum = max(maxSum, node->val + [1]);
    return node->val + max(left, right);
}

Drag options to blanks, or click blank then click option'

Aleft * right

Bmax(left, right)

Cleft + right

Dmin(left, right)

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of addition.

Using only one child sum instead of both.

Using min instead of max.

The maximum path sum through the current node includes the node's value plus both left and right child sums.

3fill in blank

hard

Fix the error in the base case to correctly handle null nodes.

DSA C++

int maxPathDown(TreeNode* node) {
    if (node == [1]) return 0;
    int left = max(0, maxPathDown(node->left));
    int right = max(0, maxPathDown(node->right));
    maxSum = max(maxSum, node->val + left + right);
    return node->val + max(left, right);
}

Drag options to blanks, or click blank then click option'

Anullptr

BNULL

C0

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or false instead of nullptr.

Using NULL which is legacy in modern C++.

In C++, the correct way to check for a null pointer is 'nullptr'.

4fill in blank

hard

Fill both blanks to complete the function that returns the maximum path sum in the tree.

DSA C++

int maxPathSum(TreeNode* root) {
    maxSum = INT_MIN;
    [1](root);
    return [2];
}

Drag options to blanks, or click blank then click option'

AmaxPathDown

BmaxSum

Croot

DmaxPathSum

Attempts:

3 left

💡 Hint

Common Mistakes

Returning root instead of maxSum.

Calling maxPathSum recursively instead of maxPathDown.

We call maxPathDown to compute max sums and then return the global maxSum.

5fill in blank

hard

Fill all three blanks to complete the recursive function that updates maxSum and returns the max path sum down one side.

DSA C++

int maxPathDown(TreeNode* node) {
    if (node == nullptr) return 0;
    int left = max(0, [1](node->left));
    int right = max(0, [2](node->right));
    maxSum = max(maxSum, node->val + left + right);
    return node->val + [3](left, right);
}

Drag options to blanks, or click blank then click option'

AmaxPathDown

Bmax

Cmin

DmaxSum

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max for choosing child path.

Returning maxSum instead of node value plus max child sum.

Calling wrong function names.

We recursively call maxPathDown on children, update maxSum, and return node value plus max of left and right.

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