Path Sum Root to Leaf in Binary Tree in DSA Java - 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.

public boolean hasPathSum(TreeNode root, int targetSum) {
    if (root == null) return false;
    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 if there is a root-to-leaf path where the sum of node values equals targetSum.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Each node is visited once in the worst case.

How Execution Grows With Input

As the number of nodes grows, the function checks each node once to find the path.

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

Pattern observation: The work grows roughly in direct proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check the path sum 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 by visiting every node until it finds a valid path or finishes checking all.

Interview Connect

Understanding how recursive tree traversal scales helps you explain your approach clearly and shows you grasp fundamental algorithm analysis.

Self-Check

"What if the tree is balanced versus very skewed? How would that affect the time complexity?"

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 tree

D. The sum of all node values in the tree

Solution

1. Step 1: Understand the problem statement

   The problem asks if any path from the root node down to any leaf node sums exactly to a given target value.

2. Step 2: Identify what is being checked

   We check if such a path exists, not the maximum depth or total sum of all nodes.

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 check = existence of target sum path [OK]

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

Common Mistakes:

• Confusing total tree sum with path sum
• Checking any path, not root-to-leaf
• Looking for maximum depth instead

2. Which of the following Java method signatures correctly represents a function to check path sum in a binary tree node class?

easy

A. public void hasPathSum(TreeNode root, int targetSum)

B. public int hasPathSum(TreeNode root, int targetSum)

C. public String hasPathSum(TreeNode root, int targetSum)

D. public boolean hasPathSum(TreeNode root, int targetSum)

Solution

1. Step 1: Identify the return type for path sum check

   The function should return true or false, so the return type must be boolean.

2. Step 2: Check method signature correctness

   public boolean hasPathSum(TreeNode root, int targetSum) returns boolean and takes root node and target sum as parameters, matching the problem requirements.

3. Final Answer:

   public boolean hasPathSum(TreeNode root, int targetSum) -> Option D

4. Quick Check:

   Return type boolean for true/false check [OK]

Hint: Return boolean for yes/no path sum check [OK]

Common Mistakes:

• Using int or void return type instead of boolean
• Returning String instead of boolean
• Incorrect parameter types

3. What will be the output of the following code snippet?

class TreeNode {
  int val;
  TreeNode left, right;
  TreeNode(int val) { this.val = val; }
}

public class Solution {
  public boolean hasPathSum(TreeNode root, int targetSum) {
    if (root == null) return false;
    if (root.left == null && root.right == null) return root.val == targetSum;
    return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);
  }

  public static void main(String[] args) {
    TreeNode root = new TreeNode(5);
    root.left = new TreeNode(4);
    root.right = new TreeNode(8);
    root.left.left = new TreeNode(11);
    root.left.left.left = new TreeNode(7);
    root.left.left.right = new TreeNode(2);
    root.right.left = new TreeNode(13);
    root.right.right = new TreeNode(4);
    root.right.right.right = new TreeNode(1);

    Solution sol = new Solution();
    System.out.println(sol.hasPathSum(root, 22));
  }
}

medium

A. true

B. false

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Trace the path sums in the tree

   Check if any root-to-leaf path sums to 22. One path is 5->4->11->2 = 22.

2. Step 2: Evaluate the recursive calls

   The function returns true because the path sum 22 exists, so it prints true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Path 5+4+11+2 = 22 exists [OK]

Hint: Look for any root-to-leaf path matching target sum [OK]

Common Mistakes:

• Ignoring leaf node condition
• Subtracting sum incorrectly
• Assuming sum must be from root to any node

4. Identify the bug in the following code for checking path sum and select the correct fix:

public boolean hasPathSum(TreeNode root, int targetSum) {
  if (root == null) return false;
  if (root.left == null && root.right == null) return root.val == targetSum;
  return hasPathSum(root.left, targetSum) || hasPathSum(root.right, targetSum);
}

medium

A. Change return type to int

B. Subtract root.val from targetSum in recursive calls

C. Check only left subtree for path sum

D. Return true if root is null

Solution

1. Step 1: Analyze recursive call parameters

   The code calls hasPathSum with the same targetSum without subtracting root.val, so it doesn't reduce the sum along the path.

2. Step 2: Correct the recursive calls

   Subtract root.val from targetSum in recursive calls to track remaining sum needed.

3. Final Answer:

   Subtract root.val from targetSum in recursive calls -> Option B

4. Quick Check:

   Update targetSum by subtracting node value [OK]

Hint: Always reduce targetSum by current node's value [OK]

Common Mistakes:

• Not subtracting node value in recursion
• Returning true on null node
• Ignoring right subtree

5. Given a binary tree where node values can be negative, zero, or positive, which modification is necessary to correctly check for a path sum equal to a target value?

hard

A. Check for path sums only in positive-valued nodes

B. Use a global variable to track sums instead of recursion

C. No modification needed; standard path sum check works for all values

D. Modify the function to handle negative sums by not stopping at leaf nodes

Solution

1. Step 1: Understand impact of negative values on path sum

   Negative values do not affect the logic; the sum is still accumulated along the path.

2. Step 2: Confirm standard recursive approach works

   The standard recursive subtraction and leaf check works regardless of node values being negative, zero, or positive.

3. Final Answer:

   No modification needed; standard path sum check works for all values -> Option C

4. Quick Check:

   Standard method handles all integer values [OK]

Hint: Standard recursion handles negative values fine [OK]

Common Mistakes:

• Thinking negative values break the logic
• Trying to skip negative nodes
• Changing leaf node condition unnecessarily