Why DP on Trees Maximum Path Sum in DSA C?

The Big Idea

What if you could find the strongest path in a complex tree without checking every possibility?

The Scenario

Imagine you have a family tree, and you want to find the path through relatives that gives the highest combined happiness score. Doing this by checking every possible path by hand is like trying to find a needle in a haystack.

The Problem

Manually checking every path in a tree is slow and confusing because the number of paths grows very fast. You might miss some paths or repeat work, making it easy to make mistakes and waste time.

The Solution

Dynamic Programming on Trees breaks the problem into smaller parts, solving each once and reusing the answers. This way, you quickly find the maximum path sum without checking every path again and again.

Before vs After

✗ Before

int maxPathSum(Node* root) {
    // Try all paths manually - very complex and slow
    // No reuse of results
}

✓ After

int maxPathSumUtil(Node* 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->value);
    return node->value + (left > right ? left : right);
}

What It Enables

This method lets you find the best path sum in a tree efficiently, even when the tree is very large.

Real Life Example

Finding the strongest connection path in a social network or the most profitable route in a company's organizational chart.

Key Takeaways

Manual checking of all paths is slow and error-prone.

DP on Trees solves subproblems once and reuses results.

It efficiently finds the maximum path sum in a tree.

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 along any path between two nodes in the tree

B. The number of nodes in the longest path of the tree

C. The sum of all node values in the tree

D. The height of the tree

Solution

1. Step 1: Understand the problem definition

   The maximum path sum is about finding the path between any two nodes that yields the highest sum of node values.

2. Step 2: Differentiate from other tree properties

   Unlike longest path or height, maximum path sum focuses on the sum of node values, not count or depth.

3. Final Answer:

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

4. Quick Check:

   Maximum path sum = largest node value sum [OK]

Hint: Maximum path sum is about sums, not counts or heights [OK]

Common Mistakes:

• Confusing maximum path sum with longest path length
• Thinking it is the sum of all nodes
• Mixing it up with tree height

2. Which of the following is the correct base case initialization for a recursive function computing maximum path sum in a binary tree node in C?

easy

A. if (node == NULL) return INT_MIN;

B. if (node == NULL) return -1;

C. if (node == NULL) return 1;

D. if (node == NULL) return 0;

Solution

1. Step 1: Understand base case for null nodes

   For maximum path sum, null nodes should contribute 0 to the path sum as they do not add any value.

2. Step 2: Choose appropriate return value

   Returning 0 ensures that null nodes do not affect the sum and allows the recursion to proceed correctly.

3. Final Answer:

   if (node == NULL) return 0; -> Option D

4. Quick Check:

   Base case returns 0 to represent no contribution from null nodes [OK]

Hint: Use 0 for null nodes to avoid affecting sums [OK]

Common Mistakes:

• Returning INT_MIN which can cause incorrect sums
• Returning positive values for null nodes
• Returning 1 or -1 which are arbitrary and incorrect

3. Consider the following C code snippet for maximum path sum in a binary tree node structure. What is the output printed by printf("%d", maxSum); after calling maxPathSum(root); on the tree below?

Tree structure:
1
/ \\
2 3

typedef struct Node {
    int val;
    struct Node* left;
    struct Node* right;
} Node;

int maxSum = INT_MIN;

int maxPathSumUtil(Node* root) {
    if (!root) return 0;
    int left = maxPathSumUtil(root->left);
    int right = maxPathSumUtil(root->right);
    int maxSingle = (left > right ? left : right) + root->val;
    int maxTop = (left + right + root->val);
    if (maxTop > maxSum) maxSum = maxTop;
    return maxSingle > root->val ? maxSingle : root->val;
}

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

medium

A. 3

B. 5

C. 6

D. 4

Solution

1. Step 1: Calculate max path sums for each node

   For node 2 and 3, maxSingle = 2 and 3 respectively, maxTop = 2 and 3. For root 1, left=2, right=3, maxSingle = max(2,3)+1=4, maxTop=2+3+1=6.

2. Step 2: Update maxSum and return final value

   maxSum updates to 6 at root. The function returns maxSum = 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Sum of path 2->1->3 = 6 [OK]

Hint: Sum path through root and children for max sum [OK]

Common Mistakes:

• Ignoring path through both children
• Returning maxSingle instead of maxSum
• Not updating global maxSum correctly

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

int maxSum = INT_MIN;

int maxPathSumUtil(Node* root) {
    if (!root) return 0;
    int left = maxPathSumUtil(root->left);
    int right = maxPathSumUtil(root->right);
    int maxSingle = (left > right ? left : right) + root->val;
    int maxTop = (left + right + root->val);
    if (maxSingle > maxSum) maxSum = maxSingle;
    if (maxTop > maxSum) maxSum = maxTop;
    return maxSingle;
}

medium

A. It returns maxSingle without comparing it to root->val

B. It updates maxSum with maxSingle before maxTop, missing some max sums

C. It does not handle negative node values correctly

D. It returns maxSingle but should return maxTop

Solution

1. Step 1: Check return value correctness

   maxSingle should be max of (maxSingle, root->val) to handle negative values properly.

2. Step 2: Identify missing max comparison

   Without comparing maxSingle to root->val, the function may return a smaller sum than root's value alone.

3. Final Answer:

   It returns maxSingle without comparing it to root->val -> Option A

4. Quick Check:

   Return value must consider root->val for negatives [OK]

Hint: Always compare maxSingle with root->val before returning [OK]

Common Mistakes:

• Not handling negative values correctly
• Updating maxSum incorrectly
• Returning maxTop instead of maxSingle

5. Given a binary tree where nodes can have negative values, which approach correctly computes the maximum path sum using dynamic programming on trees?

hard

A. Use a recursive function that returns the maximum path sum including the current node and one child, update a global max with sum of node and both children, and handle negative values by comparing with zero

B. Use a recursive function that returns the maximum path sum including the current node and one child, update a global max with sum of node and both children, and handle negative values by comparing with the node's value itself

C. Use a recursive function that returns the sum of all nodes in the subtree, ignoring negative values, and update global max accordingly

D. Use a recursive function that returns the maximum path sum including the current node and both children always, without any comparison for negative values

Solution

1. Step 1: Understand handling of negative values

   Negative values require comparing max sums with the node's own value to avoid reducing sums by adding negative children.

2. Step 2: Correct recursive approach

   Return max path sum including current node and one child or node alone; update global max with sum of node and both children.

3. Final Answer:

   Use a recursive function that returns the maximum path sum including the current node and one child, update a global max with sum of node and both children, and handle negative values by comparing with the node's value itself -> Option B

4. Quick Check:

   Compare with node value to handle negatives correctly [OK]

Hint: Compare sums with node value itself to handle negatives [OK]

Common Mistakes:

• Ignoring negative values in sums
• Always adding both children without max checks
• Ignoring node's own value in comparisons