Why Path Sum Root to Leaf in Binary Tree in DSA Java?

The Big Idea

What if you could instantly know if a path in a tree adds up to your target without checking every step yourself?

The Scenario

Imagine you have a family tree drawn on paper, and you want to find if there is a path from the oldest ancestor to a youngest family member where the ages add up to a certain number. Doing this by checking every possible path manually is tiring and confusing.

The Problem

Manually adding ages along every path is slow and easy to make mistakes. You might miss some paths or add wrong numbers. It becomes impossible when the tree is big or has many branches.

The Solution

Using the "Path Sum Root to Leaf" method, a computer can quickly check each path from the root to leaves and add the values automatically. It stops early if the sum is too big or finds the correct path without checking all paths.

Before vs After

✗ Before

int sum = 0;
// Manually add values for each path
if (root != null) {
  sum += root.val;
  if (root.left != null) sum += root.left.val;
  if (root.right != null) sum += root.right.val;
}
// Check if sum equals target

✓ After

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);
}

What It Enables

This concept lets us quickly find if any root-to-leaf path adds up to a target sum, even in large trees, without checking every path by hand.

Real Life Example

In a decision tree for a game, you might want to know if there is a sequence of moves that leads to a certain score. Path Sum helps find if such a sequence exists.

Key Takeaways

Manually checking sums on tree paths is slow and error-prone.

Path Sum Root to Leaf automates checking sums along paths efficiently.

It helps find if a path with a specific sum exists in a tree quickly.

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