Maximum Path Sum in Binary Tree in DSA C++ - Execution Trace

Concept Flow - Maximum Path Sum in Binary Tree

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 sum 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 max path sum stored globally

The algorithm visits each node, calculates max path sums from left and right children, updates the global max path sum including the current node, and returns the max sum path extending to one child.

Execution Sample

DSA C++

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

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

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

Execution Table

Step	Operation	Node Visited	Left Gain	Right Gain	Price New Path	Global Max Updated	Return Value	Visual State
1	Visit node	1 (root)	?	?	?	INT_MIN	?	1
2	Visit left child	2	?	?	?	INT_MIN	?	1 -> 2
3	Visit left child of 2	4	0	0	4	4	4	1 -> 2 -> 4
4	Return from 4	4	0	0	4	4	4	1 -> 2 -> 4
5	Visit right child of 2	5	0	0	5	5	5	1 -> 2 -> 4,5
6	Return from 5	5	0	0	5	5	5	1 -> 2 -> 4,5
7	Calculate at 2	2	4	5	11	11	7	1 -> 2 -> 4,5
8	Return from 2	2	4	5	11	11	7	1 -> 2 -> 4,5
9	Visit right child	3	?	?	?	11	?	1 -> 2 -> 4,5,3
10	Visit left child of 3	6	0	0	6	11	6	1 -> 2 -> 4,5,3,6
11	Return from 6	6	0	0	6	11	6	1 -> 2 -> 4,5,3,6
12	Visit right child of 3	7	0	0	7	11	7	1 -> 2 -> 4,5,3,6,7
13	Return from 7	7	0	0	7	11	7	1 -> 2 -> 4,5,3,6,7
14	Calculate at 3	3	6	7	16	16	10	1 -> 2 -> 4,5,3,6,7
15	Return from 3	3	6	7	16	16	10	1 -> 2 -> 4,5,3,6,7
16	Calculate at 1 (root)	1	7	10	18	18	11	1 -> 2 -> 4,5,3,6,7
17	Return from 1	1	7	10	18	18	11	1 -> 2 -> 4,5,3,6,7
18	End	-	-	-	-	18	-	Final max path sum = 18

💡 All nodes visited, recursion unwound, global max path sum is 18

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 7	After Step 10	After Step 12	After Step 14	After Step 16	Final
maxSum	INT_MIN	4	5	11	11	11	16	18	18
leftGain (node 2)	?	4	4	4	4	4	4	4	4
rightGain (node 2)	?	?	5	5	5	5	5	5	5
leftGain (node 3)	?	?	?	?	6	6	6	6	6
rightGain (node 3)	?	?	?	?	?	7	7	7	7
return value (node 1)	?	?	?	?	?	?	?	11	11

Key Moments - 3 Insights

Why do we take max with 0 for leftGain and rightGain?

Because negative gains would reduce the path sum, we ignore them by taking max with 0 as shown in steps 3, 5, 10, and 12 where gains are set to 0 if negative.

Why do we update the global max with priceNewPath instead of return value?

The global max considers the path through the current node including both children (step 7, 14, 16), but the return value only includes the node plus one child's max gain to maintain path continuity upwards.

What does the return value represent at each node?

It represents the maximum gain from the current node to one of its subtrees, used by the parent node to compute its path sums, as seen in steps 7, 14, and 16.

Visual Quiz - 3 Questions

Look at the execution table at step 14, what is the priceNewPath at node 3?

A 13

B 10

C 16

D 18

💡 Hint

Check the 'Price New Path' column at step 14 in the execution_table.

At which step does the global max first update to 11?

A Step 5

B Step 7

C Step 14

D Step 16

💡 Hint

Look at the 'Global Max Updated' column and find when it changes to 11.

If node 5 had a negative value, how would that affect the leftGain or rightGain at node 2?

A rightGain would be zero

B leftGain would be negative

C rightGain would be negative

D leftGain would be zero

💡 Hint

Recall that negative gains are replaced by zero as per the max with 0 rule in variable_tracker.

Concept Snapshot

Maximum Path Sum in Binary Tree:
- Visit each node recursively
- Calculate max gain from left and right children (ignore negatives)
- Update global max with node.val + leftGain + rightGain
- Return node.val + max(leftGain, rightGain) to parent
- Final global max is the answer

Full Transcript

The maximum path sum in a binary tree is found by visiting each node and calculating the maximum gain from its left and right children. Negative gains are ignored by taking the maximum with zero. At each node, the algorithm calculates the sum of the node's value plus the left and right gains, updating a global maximum if this sum is higher. The function returns the node's value plus the maximum gain from one child to maintain path continuity. This process repeats recursively for all nodes, and the global maximum path sum is returned at the end.