🧑‍💻 Interviewer Guide: Path Sum (UMPIRE Method)

https://leetcode.com/problems/path-sum/

📌 Problem Statement

Given the root of a binary tree and an integer targetSum,
Return True if the tree has a root-to-leaf path such that the sum of the node values along the path equals targetSum.

A leaf is a node with no children.

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🧭 U — Understand the Problem

Interviewer Prompts & Possible Answers:

• "Can you restate the problem in your own words?"
  → "Sure, we need to check if there's a path from the root node to any leaf where the sum of the node values equals the given target."

• "What's the difference between any path and a root-to-leaf path?"
  → "Any path could end at an internal node, but a root-to-leaf path must go all the way down to a node with no children."

• "What qualifies a node as a leaf in this context?"
  → "A node with no left or right children."

• "What should the function return if the tree is empty?"
  → "If the tree is empty, there's no path at all, so it should return False."

• "Can node values be negative?"
  → "Yes, node values can be negative, which means we can't use early termination optimizations."

• "What if the tree has only one node?"
  → "A single node is both root and leaf, so we just check if its value equals the target sum."

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🧪 M — Match with Patterns

Interviewer Prompts & Possible Answers:

• "What kind of traversal do you think fits best here?"
  → "Depth-first search makes sense since we're exploring complete paths from root to leaves."

• "Have you seen problems involving root-to-leaf paths before?"
  → "Yes, similar to problems like path sum collection, max depth, or binary tree path listing."

• "Do you think recursion or iteration would work better for this?"
  → "Recursion feels more natural because the structure of the problem mirrors the recursive tree traversal. Iteration would need a stack and extra bookkeeping."

• "What's the key insight for solving this recursively?"
  → "At each node, we reduce the problem by subtracting the current node's value from the target and recursing on children."

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🛠️ P — Plan the Solution

Interviewer Prompts & Possible Answers:

• "Can you walk me through how you'd approach this recursively?"
  → "At each node, subtract the current node's value from the target. If we reach a leaf and the remaining sum is zero, we return true."

• "What's your base case?"
  → "If the node is None (empty tree), return False. If we're at a leaf node, check if the remaining sum equals the node's value."

• "How do you reduce the problem as you move down the tree?"
  → "We keep subtracting the node's value from the target and pass the new target down to the children."

• "How would you handle the target sum at each step?"
  → "I'd track the remaining sum as I recurse down, updating it by subtracting the current node's value."

• "What happens when you have both left and right children?"
  → "We need to check both subtrees. If either subtree has a valid path, we return True."

• "Could you outline the algorithm step by step?"
  → "1. Handle null tree case. 2. Check if current node is a leaf and equals remaining target. 3. Recursively check left and right subtrees with updated target. 4. Return true if either subtree has a valid path."

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⌨️ I — Implement

Recursive Solution

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def hasPathSum(root: TreeNode, targetSum: int) -> bool:
    # Base case: empty tree
    if not root:
        return False
    
    # If we're at a leaf node, check if the current value equals targetSum
    if not root.left and not root.right:
        return root.val == targetSum
    
    # Recursively check left and right subtrees with updated sum
    remainingSum = targetSum - root.val
    return (
        hasPathSum(root.left, remainingSum) or 
        hasPathSum(root.right, remainingSum)
    )

Iterative Solution (Alternative)

def hasPathSum(root: TreeNode, targetSum: int) -> bool:
    if not root:
        return False
    
    # Stack stores tuples of (node, remaining_sum_needed)
    stack = [(root, targetSum - root.val)]
    
    while stack:
        current, remaining = stack.pop()
        
        # Check if we reached a leaf node with the correct path sum
        if not current.left and not current.right and remaining == 0:
            return True
        
        # Push children to the stack with updated remaining sums
        if current.right:
            stack.append((current.right, remaining - current.right.val))
        if current.left:
            stack.append((current.left, remaining - current.left.val))
    
    return False

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🔁 R — Review

Interviewer Prompts & Possible Answers:

• "What edge cases should we test?"
  → "An empty tree, a tree with one node, trees where no path matches, trees with negative values, and trees where multiple paths exist but only one adds up correctly."

• "How would you verify this works with different types of trees?"
  → "I'd try balanced trees, skewed trees, and trees with positive, negative, and zero values."

• "Could you trace through a simple example?"
  → "Sure, for a tree [5,4,8,11,null,13,4,7,2,null,null,null,1] with target 22: 5→4→11→2 gives us 5+4+11+2=22, so return True."

• "What happens with negative numbers?"
  → "Negative numbers prevent early termination optimizations, but our algorithm handles them correctly since we check the exact sum at leaf nodes."

Test Cases:

def test_hasPathSum():
    # Empty tree
    assert hasPathSum(None, 0) == False
    
    # Single node - match
    assert hasPathSum(TreeNode(5), 5) == True
    
    # Single node - no match
    assert hasPathSum(TreeNode(1), 5) == False
    
    # Complex tree with valid path
    root = TreeNode(5,
        left=TreeNode(4,
            left=TreeNode(11, TreeNode(7), TreeNode(2))
        ),
        right=TreeNode(8,
            left=TreeNode(13),
            right=TreeNode(4, right=TreeNode(1))
        )
    )
    assert hasPathSum(root, 22) == True  # 5+4+11+2=22
    assert hasPathSum(root, 26) == True  # 5+8+13=26
    assert hasPathSum(root, 18) == True  # 5+8+4+1=18
    assert hasPathSum(root, 100) == False

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🧹 E — Evaluate & Optimize

Interviewer Prompts & Possible Answers:

• "What's the time and space complexity of your solution?"
  → "Time is O(n) since we might visit every node in the worst case. Space is O(h) where h is the tree height due to recursion stack."

• "Could this be done iteratively?"
  → "Yes, using a stack to simulate recursion and explicitly tracking the current sum along with the node."

• "What are the tradeoffs between recursive and iterative approaches?"
  → "Recursion is simpler and cleaner to write, but iterative avoids potential stack overflow issues in very deep trees."

• "Can we optimize this further?"
  → "Not really in terms of time complexity - we need to potentially check all paths. The space complexity is already optimal for the recursive approach."

• "How does this compare to collecting all root-to-leaf sums first?"
  → "Our approach is more efficient because it can return early when a valid path is found, rather than computing all possible sums."

• "What if we needed to find all valid paths instead of just checking existence?"
  → "We'd need to modify the approach to collect paths instead of returning boolean, and we couldn't use early termination."