What does the 'Path Sum Root to Leaf' problem ask you to find in a binary tree?
It asks whether there is a path from the root node to any leaf node such that the sum of the node values along that path equals a given target sum.
Click to reveal answer
beginner
In the context of 'Path Sum Root to Leaf', what is a leaf node?
A leaf node is a node in a binary tree that has no children (no left or right child).
Click to reveal answer
intermediate
Why do we subtract the current node's value from the target sum when moving down the tree?
Because we want to check if the remaining sum needed to reach the target can be found in the subtree below the current node.
Click to reveal answer
intermediate
What is the base case for the recursive solution to the Path Sum problem?
When the current node is null, return false. When the current node is a leaf, check if its value equals the remaining sum.
Click to reveal answer
beginner
How does the algorithm decide to return true or false?
It returns true if any path from root to leaf sums to the target; otherwise, it returns false after checking all paths.
Click to reveal answer
What is the first step in checking for a path sum in a binary tree?
AReturn true immediately
BAdd the current node's value to the sum
CCheck if the current node is null
DCheck if the current node has children
✗ Incorrect
We first check if the current node is null to handle the base case and avoid errors.
When do we check if the current node's value equals the remaining sum?
AAfter visiting all nodes
BWhen the node has children
CAt the root node only
DWhen the node is a leaf node
✗ Incorrect
We check this at leaf nodes because only root-to-leaf paths count.
If the target sum is 22 and the root node value is 5, what sum do we look for in the subtrees?
A17
B27
C22
D5
✗ Incorrect
We subtract the root value (5) from the target sum (22) to get 17 for the subtrees.
What does it mean if the function returns true at any point?
AA valid path sum has been found
BNo path sum exists
CThe tree is empty
DThe target sum is zero
✗ Incorrect
Returning true means a root-to-leaf path with the target sum exists.
Which traversal method is commonly used to solve the Path Sum problem?
ABreadth-first search (BFS)
BDepth-first search (DFS)
CLevel order traversal
DInorder traversal only
✗ Incorrect
DFS is used to explore each path from root to leaf.
Explain how you would check if a binary tree has a root-to-leaf path with a given sum.
Think about how to reduce the problem as you go down the tree.
You got /4 concepts.
Describe the base cases needed in a recursive solution for the Path Sum problem.
Consider what happens when you reach the end of a path.
You got /2 concepts.
Practice
(1/5)
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
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.
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.
Final Answer:
If there exists a root-to-leaf path with a sum equal to a given target -> Option A
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
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.
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.
Final Answer:
if (node === null) return false; -> Option D
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
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).
Step 2: Evaluate the recursive calls
The function returns true because the path exists, so console.log prints true.
Final Answer:
true -> Option A
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
Step 1: Identify incorrect assignment in leaf check
The line 'return sum = node.val;' uses assignment '=' instead of comparison '===' which causes wrong behavior.
Step 2: Correct the comparison operator
Replace '=' with '===' to properly check if sum equals node.val at leaf nodes.
Final Answer:
Change 'sum = node.val' to 'sum === node.val' in the leaf check -> Option C
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
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.
Step 2: Confirm no cycles in binary tree
Binary trees have no cycles, so no infinite loops occur; no visited set needed.
Final Answer:
No change needed; current code works with negative values -> Option B
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
Recommended
Ready to prove it?
Pass the test and this pattern moves unknownstrongwhile earning +1 Grip toward your certificate.