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

Time Complexity: DP on Trees Maximum Path Sum

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the maximum path sum in a tree grows as the tree gets bigger.

Specifically, how does the number of steps change when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function maxPathSum(root: TreeNode | null): number {
  let maxSum = -Infinity;
  function dfs(node: TreeNode | null): number {
    if (!node) return 0;
    const left = Math.max(dfs(node.left), 0);
    const right = Math.max(dfs(node.right), 0);
    maxSum = Math.max(maxSum, node.val + left + right);
    return node.val + Math.max(left, right);
  }
  dfs(root);
  return maxSum;
}
    

This code finds the maximum sum of values along any path in a binary tree using a depth-first search.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive depth-first search (dfs) visits each node once.
• How many times: Each node is visited exactly one time during the recursion.

How Execution Grows With Input

As the number of nodes in the tree increases, the number of recursive calls grows linearly.

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

Pattern observation: The number of steps grows directly in proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum path sum grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because the function calls itself twice per node, the time is O(2^n)."

[OK] Correct: Each node is visited only once because the recursion branches do not overlap; the calls form a tree structure with n nodes, so total calls are n, not exponential.

Interview Connect

Understanding this linear time complexity helps you explain how efficient tree algorithms work and shows you can analyze recursive solutions clearly.

Self-Check

"What if the tree was very unbalanced and looked 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 sum of all node values in the tree

B. The largest sum of node values along any path between two nodes 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 about finding the path between any two nodes that yields the highest sum of node values.

2. Step 2: Differentiate from other sums

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

3. Final Answer:

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

4. Quick Check:

   Maximum path sum = max sum on any path [OK]

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

Common Mistakes:

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

2. Which of the following TypeScript function signatures correctly represents a recursive function to find maximum path sum in a binary tree node?

easy

A. function maxPathSum(node: TreeNode): void

B. function maxPathSum(node: number): number

C. function maxPathSum(): number

D. function maxPathSum(node: TreeNode | null): number

Solution

1. Step 1: Check parameter type

   The function must accept a TreeNode or null to handle empty nodes.

2. Step 2: Check return type

   It should return a number representing the max path sum from that node.

3. Final Answer:

   function maxPathSum(node: TreeNode | null): number -> Option D

4. Quick Check:

   Correct parameter and return type [OK]

Hint: Function must accept node or null and return number [OK]

Common Mistakes:

• Using wrong parameter type (e.g., number instead of TreeNode)
• Returning void instead of number
• Not allowing null for leaf children

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

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val = 0, left: TreeNode | null = null, right: TreeNode | null = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function maxPathSum(root: TreeNode | null): number {
  let maxSum = -Infinity;
  function dfs(node: TreeNode | null): number {
    if (!node) return 0;
    const left = Math.max(dfs(node.left), 0);
    const right = Math.max(dfs(node.right), 0);
    maxSum = Math.max(maxSum, node.val + left + right);
    return node.val + Math.max(left, right);
  }
  dfs(root);
  return maxSum;
}

const tree = new TreeNode(1, new TreeNode(2), new TreeNode(3));
console.log(maxPathSum(tree));

What is the output of this code?

medium

A. 3

B. 5

C. 6

D. 1

Solution

1. Step 1: Trace DFS on tree nodes

   For node 2 and 3, max path sums are 2 and 3 respectively. For root 1, left=2, right=3.

2. Step 2: Calculate maxSum updates

   maxSum updates to max(-Infinity, 1+2+3) = 6. Return value for root is 1 + max(2,3) = 4 but maxSum remains 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

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

Hint: Sum left + node + right for max path sum [OK]

Common Mistakes:

• Returning max path from root only, ignoring subtrees
• Not using Math.max with 0 to ignore negative sums
• Confusing return value with maxSum global

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

function maxPathSum(root: TreeNode | null): number {
  let maxSum = -Infinity;
  function dfs(node: TreeNode | null): number {
    if (!node) return 0;
    const left = dfs(node.left);
    const right = dfs(node.right);
    maxSum = Math.max(maxSum, node.val + left + right);
    return node.val + Math.max(left, right);
  }
  dfs(root);
  return maxSum;
}

What is the bug in this code?

medium

A. It does not ignore negative sums from left or right subtree

B. It returns void instead of number

C. It does not handle null root

D. It uses global variable incorrectly

Solution

1. Step 1: Analyze subtree sums usage

   The code uses left and right sums directly without ignoring negative values.

2. Step 2: Identify impact on maxSum

   Negative sums can reduce maxSum, so we must use Math.max(dfs(...), 0) to ignore negatives.

3. Final Answer:

   It does not ignore negative sums from left or right subtree -> Option A

4. Quick Check:

   Negative subtree sums must be ignored [OK]

Hint: Use Math.max(dfs(...), 0) to ignore negatives [OK]

Common Mistakes:

• Not checking for negative subtree sums
• Assuming dfs returns void
• Ignoring null root case

5. Given a binary tree where some nodes have negative values, how would you modify the dynamic programming approach to find the maximum path sum that can start and end at any node, but the path must go downwards (only from parent to child)?

hard

A. Use DFS to compute max downward path sum at each node ignoring negative sums by max(0, childSum)

B. Use BFS to find the longest path ignoring node values

C. Sum all node values and subtract minimum node value

D. Only consider paths that include the root node

Solution

1. Step 1: Understand downward path constraint

   The path must go downwards from parent to child, so use DFS to explore children.

2. Step 2: Handle negative values

   Ignore negative sums by taking max(0, childSum) to avoid reducing path sum.

3. Step 3: Calculate max path sum

   At each node, max downward path sum = node.val + max(leftChildSum, rightChildSum, 0).

4. Final Answer:

   Use DFS to compute max downward path sum at each node ignoring negative sums by max(0, childSum) -> Option A

5. Quick Check:

   Downward path + ignore negatives = DFS with max(0, childSum) [OK]

Hint: Downward path max sum uses DFS and max(0, childSum) [OK]

Common Mistakes:

• Using BFS instead of DFS
• Ignoring negative values incorrectly
• Forcing path through root only