Path Sum Root to Leaf in Binary Tree in DSA C - 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 to check if a path sum exists grows as the tree gets bigger.

How does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Check if any root-to-leaf path sums to target
int hasPathSum(struct TreeNode* root, int targetSum) {
    if (root == NULL) return 0;
    if (root->left == NULL && root->right == NULL) {
        return (targetSum == root->val);
    }
    return hasPathSum(root->left, targetSum - root->val) ||
           hasPathSum(root->right, targetSum - root->val);
}
    

This code checks every path from the root to leaves to find if any path sums to the target.

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 calls
100	About 100 calls
1000	About 1000 calls

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 must check all root-to-leaf paths, so it visits every node, making the time grow with tree size.

Interview Connect

Understanding this helps you explain how recursive tree problems scale and shows you can analyze traversal costs clearly.

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 leaf 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 max depth, counting leaves, or summing all nodes. It specifically checks for a path with a certain sum.

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 means root-to-leaf sum check [OK]

Hint: Look for root-to-leaf path sums, not total or max values [OK]

Common Mistakes:

• Confusing path sum with total tree sum
• Checking any path, not root-to-leaf
• Counting nodes instead of summing values

2. Which of the following is the correct C function signature to check if a binary tree has a root-to-leaf path with a given sum?

easy

A. int hasPathSum(struct TreeNode* root, int sum);

B. void hasPathSum(struct TreeNode* root, int sum);

C. int hasPathSum(int root, int sum);

D. bool hasPathSum(struct TreeNode* root);

Solution

1. Step 1: Identify correct parameter types

   The function needs the root pointer of the tree (struct TreeNode*) and the target sum (int).

2. Step 2: Identify correct return type

   The function returns an int (usually 0 or 1) indicating false or true for path existence.

3. Final Answer:

   int hasPathSum(struct TreeNode* root, int sum); -> Option A

4. Quick Check:

   Function returns int and takes root pointer and sum [OK]

Hint: Function returns int and takes root pointer plus sum [OK]

Common Mistakes:

• Using void return type instead of int
• Passing root as int instead of pointer
• Missing sum parameter

3. Given the binary tree below and target sum = 22, what will the function hasPathSum return?

    5
   / \
  4   8
 /   / \
11  13  4
/  \      \
7    2      1

medium

A. 0

B. 1

C. Error

D. Undefined

Solution

1. Step 1: Find root-to-leaf paths and their sums

   Paths and sums: - 5->4->11->7 = 27 - 5->4->11->2 = 22 - 5->8->13 = 26 - 5->8->4->1 = 18

2. Step 2: Check if any path sum equals 22

   The path 5->4->11->2 sums to 22, so the function returns true (1).

3. Final Answer:

   1 -> Option B

4. Quick Check:

   Path sum 22 exists = 1 [OK]

Hint: Check all root-to-leaf sums carefully [OK]

Common Mistakes:

• Ignoring leaf node condition
• Stopping at non-leaf nodes
• Miscomputing sums

4. Consider this buggy C code snippet for path sum check:

int hasPathSum(struct TreeNode* root, int sum) {
  if (!root) return 0;
  if (!root->left && !root->right && root->val == sum) return 1;
  return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
}

What is the bug in this code?

medium

A. It misses subtracting root->val from sum

B. It incorrectly returns 1 for non-leaf nodes

C. It does not handle empty tree correctly

D. It correctly implements the path sum check

Solution

1. Step 1: Analyze base cases

   Returns 0 if root is NULL (empty tree). Returns 1 if leaf node value equals sum.

2. Step 2: Analyze recursive calls

   Recursively checks left and right subtrees with updated sum (sum - root->val).

3. Final Answer:

   It correctly implements the path sum check -> Option D

4. Quick Check:

   Code matches standard path sum logic [OK]

Hint: Check base cases and recursive calls carefully [OK]

Common Mistakes:

• Missing leaf node check
• Not subtracting root value from sum
• Returning wrong values for NULL nodes

5. You want to modify the path sum function to return all root-to-leaf paths where the sum equals the target. Which data structure is best to store the current path during recursion?

hard

A. A queue to process nodes level by level

B. A hash map to store node values and counts

C. A stack or dynamic array to track nodes in current path

D. A linked list to store all leaf nodes

Solution

1. Step 1: Understand path tracking during recursion

   To record the current path from root to current node, a stack or dynamic array is used to push and pop nodes as recursion goes deeper or backtracks.

2. Step 2: Differentiate from other data structures

   A queue is for breadth-first traversal, hash map is for counting, linked list for storing nodes but not efficient for backtracking paths.

3. Final Answer:

   A stack or dynamic array to track nodes in current path -> Option C

4. Quick Check:

   Use stack/array for path tracking [OK]

Hint: Use stack or array to track path during recursion [OK]

Common Mistakes:

• Using queue instead of stack for path tracking
• Not backtracking path correctly
• Using hash map for path storage