Path Sum Root to Leaf in Binary Tree in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Rust code for path sum check?

Given the binary tree and target sum, what does the function return?

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct TreeNode {
    val: i32,
    left: Option<Rc<RefCell<TreeNode>>>,
    right: Option<Rc<RefCell<TreeNode>>>,
}

impl TreeNode {
    fn new(val: i32) -> Rc<RefCell<TreeNode>> {
        Rc::new(RefCell::new(TreeNode { val, left: None, right: None }))
    }
}

fn has_path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> bool {
    match root {
        None => false,
        Some(node) => {
            let node = node.borrow();
            if node.left.is_none() && node.right.is_none() {
                return node.val == target_sum;
            }
            has_path_sum(node.left.clone(), target_sum - node.val) || has_path_sum(node.right.clone(), target_sum - node.val)
        }
    }
}

fn main() {
    let root = TreeNode::new(5);
    root.borrow_mut().left = Some(TreeNode::new(4));
    root.borrow_mut().right = Some(TreeNode::new(8));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(11));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(13));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(4));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(7));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(2));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().right.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(1));

    let result = has_path_sum(Some(root), 22);
    println!("{}", result);
}

A true

B false

C Compilation error

D Runtime panic

💡 Hint

Check if any root-to-leaf path sums exactly to 22.

The path 5->4->11->2 sums to 22, so the function returns true.

❓ Predict Output

intermediate

What is the output of this Rust code for path sum check with no valid path?

Given the binary tree and target sum, what does the function return?

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct TreeNode {
    val: i32,
    left: Option<Rc<RefCell<TreeNode>>>,
    right: Option<Rc<RefCell<TreeNode>>>,
}

impl TreeNode {
    fn new(val: i32) -> Rc<RefCell<TreeNode>> {
        Rc::new(RefCell::new(TreeNode { val, left: None, right: None }))
    }
}

fn has_path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> bool {
    match root {
        None => false,
        Some(node) => {
            let node = node.borrow();
            if node.left.is_none() && node.right.is_none() {
                return node.val == target_sum;
            }
            has_path_sum(node.left.clone(), target_sum - node.val) || has_path_sum(node.right.clone(), target_sum - node.val)
        }
    }
}

fn main() {
    let root = TreeNode::new(1);
    root.borrow_mut().left = Some(TreeNode::new(2));
    root.borrow_mut().right = Some(TreeNode::new(3));

    let result = has_path_sum(Some(root), 5);
    println!("{}", result);
}

A Compilation error

B true

C Runtime panic

D false

💡 Hint

Check if any root-to-leaf path sums exactly to 5.

No root-to-leaf path sums to 5 (paths are 1->2=3 and 1->3=4).

🔧 Debug

advanced

What error does this Rust code produce when checking path sum?

Identify the error when running this code snippet.

DSA Rust

fn has_path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> bool {
    if root.is_none() {
        return false;
    }
    let node = root.unwrap().borrow();
    if node.left.is_none() && node.right.is_none() {
        return node.val == target_sum;
    }
    has_path_sum(node.left.clone(), target_sum - node.val) || has_path_sum(node.right.clone(), target_sum - node.val)
}

A Compilation error: missing lifetime specifier

B Compilation error: cannot move out of borrowed content

C No error, runs correctly

D Runtime panic: unwrap on None value

💡 Hint

Check how unwrap is used on borrowed data.

Calling unwrap() on root moves the value, but root is borrowed, causing a move error.

🧠 Conceptual

advanced

How does the path sum function handle leaf nodes in a binary tree?

In the path sum algorithm, what condition identifies a leaf node and what check is performed?

A A leaf node has no children; check if node value is greater than remaining sum

B A leaf node has one child; check if node value is less than remaining sum

C A leaf node has no children; check if node value equals remaining sum

D A leaf node has two children; check if node value equals remaining sum

💡 Hint

Leaf nodes have no left or right children.

Leaf nodes are identified by no left and right children; the function checks if node value equals the target sum at that point.

🚀 Application

expert

What is the number of root-to-leaf paths with sum 15 in this tree?

Given the tree below, how many root-to-leaf paths sum to 15? Tree structure: - Root 5 - Left 4 - Left 11 - Left 7 - Right 2 - Right 8 - Left 13 - Right 4 - Right 1

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct TreeNode {
    val: i32,
    left: Option<Rc<RefCell<TreeNode>>>,
    right: Option<Rc<RefCell<TreeNode>>>,
}

impl TreeNode {
    fn new(val: i32) -> Rc<RefCell<TreeNode>> {
        Rc::new(RefCell::new(TreeNode { val, left: None, right: None }))
    }
}

fn count_path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> i32 {
    match root {
        None => 0,
        Some(node) => {
            let node = node.borrow();
            if node.left.is_none() && node.right.is_none() {
                return if node.val == target_sum { 1 } else { 0 };
            }
            count_path_sum(node.left.clone(), target_sum - node.val) + count_path_sum(node.right.clone(), target_sum - node.val)
        }
    }
}

fn main() {
    let root = TreeNode::new(5);
    root.borrow_mut().left = Some(TreeNode::new(4));
    root.borrow_mut().right = Some(TreeNode::new(8));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(11));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(13));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(4));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left.as_ref().unwrap().borrow_mut().left = Some(TreeNode::new(7));
    root.borrow_mut().left.as_ref().unwrap().borrow_mut().left.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(2));
    root.borrow_mut().right.as_ref().unwrap().borrow_mut().right.as_ref().unwrap().borrow_mut().right = Some(TreeNode::new(1));

    let count = count_path_sum(Some(root), 15);
    println!("{}", count);
}

A 0

B 2

C 3

D 1

💡 Hint

Count all root-to-leaf paths where sum equals 15.

No root-to-leaf path sums to 15 in the given tree. The paths and their sums are: - 5->4->11->7 = 27 - 5->4->11->2 = 22 - 5->8->13 = 26 - 5->8->4->1 = 18 Thus, the count is 0.

Practice

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

easy

A. The maximum depth of the binary tree

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

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 the maximum depth, counting nodes, or summing all nodes. It specifically checks for a path sum.

3. Final Answer:

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

4. Quick Check:

   Path sum means root-to-leaf sum check = A [OK]

Hint: Focus on root-to-leaf paths and their sums [OK]

Common Mistakes:

• Confusing path sum with total tree sum
• Checking any path, not just root-to-leaf
• Mixing depth or node count with sum

2. Which of the following Rust function signatures correctly represents a function to check path sum in a binary tree node defined as struct TreeNode { val: i32, left: Option, right: Option }?

easy

A. fn has_path_sum(root: Option<TreeNode>, sum: i32) -> bool

B. fn has_path_sum(root: TreeNode, sum: i32) -> i32

C. fn has_path_sum(root: Option, sum: i32) -> bool

D. fn has_path_sum(root: &TreeNode, sum: i32) -> Option<i32>

Solution

1. Step 1: Analyze the tree node ownership and option usage

   The tree nodes are wrapped in Option to allow for empty children and heap allocation. The function should accept Option to handle empty trees.

2. Step 2: Check return type and parameters

   The function returns a boolean indicating if a path sum exists. So the signature fn has_path_sum(root: Option, sum: i32) -> bool matches the problem.

3. Final Answer:

   fn has_path_sum(root: Option, sum: i32) -> bool -> Option C

4. Quick Check:

   Correct ownership and return type = D [OK]

Hint: Use Option for nullable tree nodes [OK]

Common Mistakes:

• Using TreeNode directly without Option or Box
• Wrong return type (not bool)
• Passing references incorrectly

3. Given the following Rust code, what is the output?

struct TreeNode {
    val: i32,
    left: Option,
    right: Option,
}

fn has_path_sum(root: Option, sum: i32) -> bool {
    match root {
        None => false,
        Some(node) => {
            if node.left.is_none() && node.right.is_none() {
                return node.val == sum;
            }
            has_path_sum(node.left, sum - node.val) || has_path_sum(node.right, sum - node.val)
        }
    }
}

fn main() {
    let leaf = TreeNode { val: 5, left: None, right: None };
    let root = TreeNode {
        val: 4,
        left: Some(Box::new(leaf)),
        right: None,
    };
    println!("{}", has_path_sum(Some(Box::new(root)), 9));
}

medium

A. Compilation error

B. false

C. Runtime panic

D. true

Solution

1. Step 1: Trace the path sum calculation

   The tree has root 4 with left child 5. The target sum is 9. The path 4 -> 5 sums to 9, so the function returns true.

2. Step 2: Confirm leaf node check and recursion

   At leaf node 5, it checks if 5 == 9 - 4 = 5, which is true, so returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   4 + 5 = 9 matches sum = true [OK]

Hint: Check leaf node sums carefully [OK]

Common Mistakes:

• Ignoring leaf node condition
• Subtracting sum incorrectly
• Confusing Option handling

4. The following Rust function is intended to check if a path sum exists but has a bug. What is the bug?

fn has_path_sum(root: Option, sum: i32) -> bool {
    if root.is_none() {
        return false;
    }
    let node = root.unwrap();
    if node.left.is_none() && node.right.is_none() {
        return node.val == sum;
    }
    has_path_sum(node.left, sum - node.val) && has_path_sum(node.right, sum - node.val)
}

medium

A. Using && instead of || for recursive calls

B. Unwrapping Option without checking None

C. Not handling empty tree case

D. Incorrect sum subtraction

Solution

1. Step 1: Analyze recursive call logic

   The function uses logical AND (&&) to combine left and right subtree results, but the problem requires checking if any path exists, so logical OR (||) is correct.

2. Step 2: Understand impact of && vs ||

   Using && means both left and right paths must have the sum, which is incorrect. Only one path needs to match.

3. Final Answer:

   Using && instead of || for recursive calls -> Option A

4. Quick Check:

   Path sum needs OR (||), not AND (&&) [OK]

Hint: Use || to check any path, not all paths [OK]

Common Mistakes:

• Using && instead of ||
• Unwrapping Option without safe check
• Ignoring leaf node condition

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

hard

A. Use backtracking with a mutable vector to store current path and a vector of vectors for results

B. Use a global variable to store paths and update it in recursion

C. Return a boolean and print paths inside the function

D. Use iterative BFS traversal without recursion

Solution

1. Step 1: Understand the requirement for all paths

   To find all root-to-leaf paths with a given sum, we need to explore all paths and store those matching the sum.

2. Step 2: Choose backtracking with mutable vectors

   Backtracking allows building paths step-by-step, storing current path in a mutable vector, and adding copies to results when sum matches.

3. Final Answer:

   Use backtracking with a mutable vector to store current path and a vector of vectors for results -> Option A

4. Quick Check:

   Backtracking with path storage = B [OK]

Hint: Backtrack paths, store copies when sum matches [OK]

Common Mistakes:

• Using global variables (not idiomatic in Rust)
• Printing instead of returning paths
• Trying BFS which is complex for path tracking