Path Sum Root to Leaf in Binary Tree in DSA Typescript - Time & Space Complexity

Time Complexity: Path Sum Root to Leaf in Binary Tree

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a path from the root to any leaf adds up to a target sum in a binary tree.

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function hasPathSum(root: TreeNode | null, targetSum: number): boolean {
  if (!root) return false;
  if (!root.left && !root.right) return root.val === targetSum;
  return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);
}
    

This code checks if there is any root-to-leaf path where the sum of node values equals the target sum.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Each node is visited once in the worst case.

How Execution Grows With Input

As the number of nodes grows, the function checks each node once to find paths.

Input Size (n)	Approx. Operations
10	About 10 node visits
100	About 100 node visits
1000	About 1000 node visits

Pattern observation: The operations grow roughly in direct proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The function only checks one path, so it runs in constant time."

[OK] Correct: The function explores all possible paths to leaves, so it must visit every node in the worst case.

Interview Connect

Understanding this helps you explain how recursive tree traversal works and how to analyze its cost, a common skill in interviews.

Self-Check

"What if the tree was a linked list (each node has only one child)? How would the time complexity change?"

Practice

1. What does the "Path Sum Root to Leaf" problem ask you to find in a binary tree?

easy

A. The sum of all node values in the binary tree

B. If there exists a root-to-leaf path with a sum equal to a given target

C. The total number of nodes in the binary tree

D. The maximum depth of the binary tree

Solution

1. Step 1: Understand the problem statement

   The problem asks if there is any path from the root node down to any leaf node where the sum of the node values equals the target sum.

2. Step 2: Differentiate from other tree properties

   It is not about depth, total nodes, or sum of all nodes, but specifically about a path sum from root to leaf.

3. Final Answer:

   If there exists a root-to-leaf path with a sum equal to a given target -> Option B

4. Quick Check:

   Path sum means root-to-leaf sum = D [OK]

Hint: Focus on root-to-leaf paths, not entire tree sums [OK]

Common Mistakes:

• Confusing path sum with total tree sum
• Checking any path, not just root-to-leaf
• Ignoring leaf node condition

2. Which of the following TypeScript function signatures correctly represents a function to check path sum in a binary tree node defined as class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; }?

easy

A. function hasPathSum(root: number, sum: TreeNode): boolean

B. function hasPathSum(root: TreeNode, sum: string): boolean

C. function hasPathSum(root: TreeNode | null, sum: number): boolean

D. function hasPathSum(root: TreeNode | null): number

Solution

1. Step 1: Check parameter types

   The root should be a TreeNode or null (empty tree), and sum should be a number representing the target sum.

2. Step 2: Check return type

   The function returns a boolean indicating if such a path exists.

3. Final Answer:

   function hasPathSum(root: TreeNode | null, sum: number): boolean -> Option C

4. Quick Check:

   Correct types and return boolean = B [OK]

Hint: Root can be null; sum is number; return boolean [OK]

Common Mistakes:

• Swapping parameter types
• Using wrong return type
• Missing null for root parameter

3. Given the following binary tree and target sum = 22, what will the function hasPathSum(root, 22) return?

    5
   / \
  4   8
 /   / \
11  13  4
/  \      
7    2     1

medium

A. null

B. false

C. SyntaxError

D. true

Solution

1. Step 1: Identify root-to-leaf paths and their sums

   Check paths: 5->4->11->7 = 27, 5->4->11->2 = 22, 5->8->13 = 26, 5->8->4->1 = 18.

2. Step 2: Check if any path sum equals 22

   Path 5->4->11->2 sums to 22, so function returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Path sum 22 exists = true [OK]

Hint: Sum each root-to-leaf path, check for target [OK]

Common Mistakes:

• Ignoring leaf node condition
• Calculating sum incorrectly
• Confusing partial paths with full root-to-leaf

4. Consider this TypeScript code snippet for checking path sum:

function hasPathSum(root: TreeNode | null, sum: number): boolean {
  if (!root) return false;
  if (!root.left && !root.right) return root.val === sum;
  return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}

What does the function return if the tree is empty (root = null) and sum is 0?

medium

A. Returns false for empty tree and sum 0

B. Causes infinite recursion

C. Throws runtime error due to null access

D. Returns true incorrectly for non-empty tree

Solution

1. Step 1: Analyze base case for empty tree

   When root is null, if (!root) returns false immediately.

2. Step 2: Confirm for sum = 0

   Empty tree has no root-to-leaf path regardless of target sum, so false is correct.

3. Final Answer:

   Returns false for empty tree and sum 0 -> Option A

4. Quick Check:

   Null root always returns false [OK]

Hint: Empty tree always returns false [OK]

Common Mistakes:

• Thinking empty tree with sum 0 should return true
• Expecting runtime error without null check
• Confusing empty tree with single-node tree

5. You want to modify the path sum function to return all root-to-leaf paths where the sum equals the target. Which approach best solves this problem?

hard

A. Use DFS to track current path and sum, collect paths when sum matches at leaf

B. Use BFS and stop at first matching path

C. Use a queue to store sums only, ignoring paths

D. Return true or false immediately when sum matches any node

Solution

1. Step 1: Understand requirement to collect all matching paths

   We need to find all root-to-leaf paths whose sums equal the target, not just check existence.

2. Step 2: Choose traversal and tracking method

   DFS (depth-first search) allows tracking the current path and sum, and collecting paths when leaf and sum match.

3. Step 3: Evaluate other options

   BFS is less natural for path tracking; queues without paths lose information; returning boolean stops early.

4. Final Answer:

   Use DFS to track current path and sum, collect paths when sum matches at leaf -> Option A

5. Quick Check:

   DFS with path tracking = A [OK]

Hint: Track path and sum with DFS to collect all matching paths [OK]

Common Mistakes:

• Using BFS without path tracking
• Stopping at first match instead of collecting all
• Ignoring leaf node condition