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

Time Complexity: Path Sum Root to Leaf in Binary Tree

O(n)

Understanding Time Complexity

We want to know how the time needed to check if a path sum exists in a binary tree grows as the tree gets bigger.

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 hasPathSum(node, targetSum) {
  if (!node) return false;
  if (!node.left && !node.right) return targetSum === node.val;
  return hasPathSum(node.left, targetSum - node.val) || hasPathSum(node.right, targetSum - node.val);
}
    

This code checks if there is a root-to-leaf path in a binary tree 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, so the steps grow roughly the same as the number of nodes.

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

Pattern observation: The work grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check the path sum grows directly 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 root-to-leaf paths, so it must visit many nodes, not just one path.

Interview Connect

Understanding this helps you explain how recursive tree problems scale, a common skill interviewers look for.

Self-Check

"What if the tree is balanced vs completely skewed? 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. If there exists a root-to-leaf path with a sum equal to a given target

B. The maximum depth of the binary tree

C. The total number of nodes in the binary tree

D. The sum of all node values in 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 a given target sum.

2. Step 2: Differentiate from other tree properties

   This is different from finding depth, counting nodes, or summing all nodes. It specifically checks for 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 A

4. Quick Check:

   Path sum problem = Check root-to-leaf sum [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 root-to-leaf
• Ignoring leaf node condition

2. Which of the following is the correct base case check in JavaScript for a null node in a recursive path sum function?

easy

A. if (node === null) return true;

B. if (node == undefined) return true;

C. if (node === undefined) return null;

D. if (node === null) return false;

Solution

1. Step 1: Identify base case for recursion

   When the current node is null, it means no path exists here, so return false to indicate failure to find the path.

2. Step 2: Check JavaScript strict equality and return value

   Using 'node === null' is correct to check null nodes, and returning false means no valid path found.

3. Final Answer:

   if (node === null) return false; -> Option D

4. Quick Check:

   Null node means no path = false [OK]

Hint: Return false on null node to stop recursion [OK]

Common Mistakes:

• Returning true on null node
• Using == instead of ===
• Returning null instead of boolean

3. What will be the output of the following code snippet?

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

function hasPathSum(node, sum) {
  if (node === null) return false;
  if (!node.left && !node.right) return sum === node.val;
  return hasPathSum(node.left, sum - node.val) || hasPathSum(node.right, sum - node.val);
}

const root = new TreeNode(5, new TreeNode(4, new TreeNode(11, new TreeNode(7), new TreeNode(2))), new TreeNode(8, null, new TreeNode(4, null, new TreeNode(1))));
console.log(hasPathSum(root, 22));

medium

A. true

B. false

C. undefined

D. TypeError

Solution

1. Step 1: Trace the path sums in the tree

   Check if any root-to-leaf path sums to 22. The path 5->4->11->2 sums to 22 (5+4+11+2=22).

2. Step 2: Evaluate the recursive calls

   The function returns true because the path exists, so console.log prints true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Path 5-4-11-2 sum=22 = true [OK]

Hint: Check leaf paths sums carefully to confirm true [OK]

Common Mistakes:

• Ignoring leaf node condition
• Miscomputing sum along path
• Assuming false when path exists

4. Find the bug in this JavaScript function for path sum and select the fix:

function hasPathSum(node, sum) {
  if (node === null) return false;
  if (!node.left && !node.right) return sum = node.val;
  return hasPathSum(node.left, sum - node.val) || hasPathSum(node.right, sum - node.val);
}

medium

A. Change 'node === null' to 'node == null' in base case

B. Remove the leaf node check entirely

C. Change 'sum = node.val' to 'sum === node.val' in the leaf check

D. Change '||' to '&&' in the return statement

Solution

1. Step 1: Identify incorrect assignment in leaf check

   The line 'return sum = node.val;' uses assignment '=' instead of comparison '===' which causes wrong behavior.

2. Step 2: Correct the comparison operator

   Replace '=' with '===' to properly check if sum equals node.val at leaf nodes.

3. Final Answer:

   Change 'sum = node.val' to 'sum === node.val' in the leaf check -> Option C

4. Quick Check:

   Use '===' for comparison, not '=' [OK]

Hint: Use '===' for equality checks, not '=' assignment [OK]

Common Mistakes:

• Using '=' instead of '===' in conditions
• Removing necessary base cases
• Changing logical operators incorrectly

5. Given a binary tree where node values can be negative, which modification is necessary to correctly check for a path sum equal to target?

function hasPathSum(node, sum) {
  if (node === null) return false;
  if (!node.left && !node.right) return sum === node.val;
  return hasPathSum(node.left, sum - node.val) || hasPathSum(node.right, sum - node.val);
}

hard

A. Use absolute values of node.val and sum in comparisons

B. No change needed; current code works with negative values

C. Change base case to return true if sum <= node.val

D. Add a visited set to avoid infinite loops with negative values

Solution

1. Step 1: Understand negative values impact

   The current recursive approach subtracts node values from sum and checks leaf equality, which works even if values are negative.

2. Step 2: Confirm no cycles in binary tree

   Binary trees have no cycles, so no infinite loops occur; no visited set needed.

3. Final Answer:

   No change needed; current code works with negative values -> Option B

4. Quick Check:

   Negative values handled by subtraction correctly [OK]

Hint: Binary trees have no cycles; negative values handled naturally [OK]

Common Mistakes:

• Adding unnecessary visited sets
• Changing base case incorrectly
• Using absolute values which breaks logic