Path Sum Root to Leaf in Binary Tree in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is null.

DSA Typescript

if (node [1] null) return false;

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

A===

B>

C!==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '===' causes wrong logic.

Using '>' or '<' operators for null check.

We check if the node is exactly equal to null to stop recursion.

2fill in blank

medium

Complete the code to check if the current node is a leaf node.

DSA Typescript

if (node.left === null && node.right [1] null) return sum + node.val === targetSum;

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

A<

B!=

C>

D===

Attempts:

3 left

💡 Hint

Common Mistakes

Using '!=' instead of '===' causes incorrect leaf detection.

Checking only one child for null.

A leaf node has both left and right children as null.

3fill in blank

hard

Fix the error in the recursive call to check left subtree.

DSA Typescript

return helper(node.left, sum [1] node.val, targetSum) || helper(node.right, sum + node.val, targetSum);

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

A/

B-

C+

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' causes decreasing sum incorrectly.

Using '*' or '/' causes wrong calculations.

We add the current node's value to the sum when going down the tree.

4fill in blank

hard

Fill both blanks to complete the function signature and initial call.

DSA Typescript

function hasPathSum([1]: TreeNode | null, targetSum: number): boolean {
  return helper([2], 0, targetSum);
}

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

Anode

Broot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'node' in parameter but calling helper with 'root'.

Mismatch between parameter name and argument.

The function parameter is named 'root' and the helper is called with 'root'.

5fill in blank

hard

Fill all three blanks to complete the helper function for path sum check.

DSA Typescript

function helper(node: TreeNode | null, sum: number, targetSum: number): boolean {
  if (node === null) return [1];
  if (node.left === null && node.right === null) return sum + node.val === [2];
  return helper(node.left, sum [3] node.val, targetSum) || helper(node.right, sum + node.val, targetSum);
}

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

Afalse

BtargetSum

C+

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Returning true when node is null causes wrong results.

Comparing sum instead of sum + node.val to targetSum.

Using wrong operator instead of '+' in recursion.

Return false if node is null, compare sum + node.val to targetSum, and add node.val to sum in recursion.

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