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

Time Complexity: Path Sum Root to Leaf in Binary Tree

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a path from the root to a leaf adds up to a target sum.

This helps us understand how the time grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code checks if any root-to-leaf path sums to the target by exploring each branch recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

As the number of nodes grows, the function visits each node once.

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

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

Final Time Complexity

Time Complexity: O(n)

This means the time to check grows linearly 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 paths until it finds a match or finishes, so it visits many nodes, not just one path.

Interview Connect

Understanding this helps you explain how recursive tree problems scale and shows you can reason about visiting all nodes efficiently.

Self-Check

"What if the tree was 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. 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