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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of path sum check for a simple tree

What is the output of the following Java code that checks if there is a root-to-leaf path with sum 22?

DSA Java

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

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

A Compilation error

B true

C false

D Runtime exception

💡 Hint

Trace the path sums from root to leaves and check if any equals 22.

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

❓ Predict Output

intermediate

Output when no path matches the sum

What does the following code print when checking for a path sum of 26?

DSA Java

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

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

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);

        Solution sol = new Solution();
        System.out.println(sol.hasPathSum(root, 26));
    }
}

A Runtime exception

B true

C Compilation error

D false

💡 Hint

Check all root-to-leaf paths and their sums.

The only paths are 1->2=3 and 1->3=4, neither equals 26, so output is false.

🔧 Debug

advanced

Identify the error causing incorrect path sum result

The following code is intended to check if a root-to-leaf path sums to a target. It always returns false. What is the bug?

DSA Java

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

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

A The leaf check uses 'sum != root.val' instead of 'sum == root.val'.

B The base case should return true when root is null.

C The method should return false if root has only one child.

D The recursive calls should use sum + root.val instead of sum - root.val.

💡 Hint

Check the condition that returns true at leaf nodes.

The leaf node check wrongly returns false when sum equals root.val, causing always false.

🧠 Conceptual

advanced

Understanding path sum with negative values

Consider a binary tree where nodes can have negative values. Which statement about the path sum algorithm is true?

A The algorithm must be changed to handle negative values correctly.

B The algorithm only works if all node values are positive.

C The existing algorithm works correctly even with negative values.

D Negative values cause infinite recursion in the algorithm.

💡 Hint

Think about how the algorithm subtracts node values from sum.

The algorithm subtracts node values from sum and checks leaf equality; negative values do not break this logic.

🚀 Application

expert

Number of root-to-leaf paths with given sum

Given the following code that counts the number of root-to-leaf paths with sum equal to target, what is the output for the tree and target 22?

DSA Java

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

public class Solution {
    public int countPathSum(TreeNode root, int sum) {
        if (root == null) return 0;
        if (root.left == null && root.right == null) return sum == root.val ? 1 : 0;
        return countPathSum(root.left, sum - root.val) + countPathSum(root.right, sum - 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.left = new TreeNode(5);
        root.right.right.right = new TreeNode(1);

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

A 2

B 3

C 1

D 4

💡 Hint

Count all root-to-leaf paths that sum to 22.

There are two paths: 5->4->11->2 and 5->8->4->5, so output is 2.

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