Path Sum Root to Leaf in Binary Tree in DSA C++ - Build from Scratch

Path Sum Root to Leaf in Binary Tree

📖 Scenario: Imagine you have a family tree represented as a binary tree. Each person has a value representing their age. You want to find if there is a path from the oldest ancestor (root) down to a youngest descendant (leaf) where the sum of ages equals a specific number.

🎯 Goal: Build a program that checks if there exists a root-to-leaf path in a binary tree such that the sum of the node values along the path equals a given target sum.

📋 What You'll Learn

Create a binary tree node structure with integer values

Build a sample binary tree with exact node values

Create a target sum variable

Write a function to check if a root-to-leaf path sums to the target

Print the result as true or false

💡 Why This Matters

🌍 Real World

Checking path sums in trees is useful in decision trees, network routing, and financial modeling where paths represent sequences of choices or transactions.

💼 Career

Understanding tree traversal and recursion is fundamental for software engineers working with hierarchical data, algorithms, and system design.

1

Create the Binary Tree Node Structure and Build the Tree

Define a struct called TreeNode with an integer val, and two pointers left and right. Then create the following binary tree exactly: root node with value 5, left child 4, right child 8, left child of 4 is 11, left and right children of 11 are 7 and 2, right child of 8 is 4, and right child of this 4 is 1.

DSA C++

// Define TreeNode struct and create the binary tree as described
// Your code here

Hint

Start by defining the TreeNode struct with a constructor. Then create each node and link them exactly as described.

2

Create the Target Sum Variable

Inside main(), create an integer variable called targetSum and set it to 22.

DSA C++

#include 
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

int main() {
    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->right = new TreeNode(4);
    root->right->right->right = new TreeNode(1);
    
    // Create an integer variable called targetSum and set it to 22
    // Your code here
    return 0;
}

Hint

Simply declare int targetSum = 22; inside main().

3

Write the Function to Check Path Sum

Write a function called hasPathSum that takes a TreeNode* called root and an integer sum. It returns true if there is a root-to-leaf path where the sum of node values equals sum, otherwise false. Use recursion and check if the current node is null, if it is a leaf, and subtract the node's value from sum as you go down.

DSA C++

#include 
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

// Write the hasPathSum function here
// Your code here

int main() {
    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->right = new TreeNode(4);
    root->right->right->right = new TreeNode(1);
    
    int targetSum = 22;
    return 0;
}

Hint

Use recursion to check if the current node is null, then if it's a leaf check if sum == root->val. Otherwise, subtract root->val from sum and check left and right subtrees.

4

Print the Result of Path Sum Check

In main(), call hasPathSum(root, targetSum) and print true if it returns true, otherwise print false.

DSA C++

#include 
using namespace std;

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

bool hasPathSum(TreeNode* root, int sum) {
    if (root == nullptr) {
        return false;
    }
    if (root->left == nullptr && root->right == nullptr) {
        return sum == root->val;
    }
    int remainingSum = sum - root->val;
    return hasPathSum(root->left, remainingSum) || hasPathSum(root->right, remainingSum);
}

int main() {
    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->right = new TreeNode(4);
    root->right->right->right = new TreeNode(1);
    
    int targetSum = 22;
    
    // Print true if hasPathSum(root, targetSum) is true, else print false
    // Your code here
    return 0;
}

Hint

Call hasPathSum(root, targetSum) and print true if it returns true, else print false.

Practice

1. What does the "Path Sum Root to Leaf" problem ask you to find in a binary tree?

easy

A. The total number of leaf nodes in the tree

B. If there exists a root-to-leaf path with a sum equal to a given target

C. The maximum depth of the binary tree

D. The sum of all node values in the 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 target sum.

2. Step 2: Differentiate from other tree properties

   It is not about depth, counting leaves, or total sum, but specifically about a path sum from root to leaf.

3. Final Answer:

   If there exists a root-to-leaf path with a sum equal to a given target -> Option B

4. Quick Check:

   Path sum means root-to-leaf sum check [OK]

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

Common Mistakes:

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

2. Which of the following C++ function signatures correctly represents a function to check path sum in a binary tree node structure?

easy

A. bool hasPathSum(TreeNode root, int targetSum)

B. int hasPathSum(TreeNode root, int targetSum)

C. void hasPathSum(TreeNode* root, int targetSum)

D. bool hasPathSum(TreeNode* root, int targetSum)

Solution

1. Step 1: Check parameter types and return type

   The function should take a pointer to the root node (TreeNode*) and an integer target sum, and return a boolean indicating if such a path exists.

2. Step 2: Validate each option

   bool hasPathSum(TreeNode* root, int targetSum) uses pointer and bool return type, which is correct. Options B and D pass TreeNode by value (less efficient), and B returns int, C returns void which is incorrect.

3. Final Answer:

   bool hasPathSum(TreeNode* root, int targetSum) -> Option D

4. Quick Check:

   Pointer input and bool output needed [OK]

Hint: Use pointer for tree nodes and bool for yes/no answers [OK]

Common Mistakes:

• Passing node by value instead of pointer
• Using wrong return type like void or int
• Missing pointer symbol (*) in parameter

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

#include <iostream>

struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

bool hasPathSum(TreeNode* root, int targetSum) {
  if (!root) return false;
  if (!root->left && !root->right) return root->val == targetSum;
  return hasPathSum(root->left, targetSum - root->val) || hasPathSum(root->right, targetSum - root->val);
}

int main() {
  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);

  bool result = hasPathSum(root, 22);
  std::cout << (result ? "true" : "false") << std::endl;
  return 0;
}

medium

A. true

B. false

C. Compilation error

D. Runtime error

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: Confirm function logic and output

   The function correctly returns true if such a path exists. The code prints "true".

3. Final Answer:

   true -> Option A

4. Quick Check:

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

Hint: Check leaf paths sums carefully [OK]

Common Mistakes:

• Ignoring leaf node condition
• Subtracting values incorrectly
• Assuming no path matches without checking all

4. Identify the bug in the following code that attempts to check path sum in a binary tree:

bool hasPathSum(TreeNode* root, int targetSum) {
  if (root == nullptr) return false;
  if (root->left == nullptr && root->right == nullptr) return targetSum == root->val;
  return hasPathSum(root->left, targetSum) || hasPathSum(root->right, targetSum);
}

medium

A. The leaf node check should return false instead of comparison

B. The base case incorrectly returns false for nullptr

C. The targetSum is not reduced by root->val in recursive calls

D. The function should return int instead of bool

Solution

1. Step 1: Analyze recursive calls

   The function calls itself with the same targetSum without subtracting the current node's value, so it never reduces the sum along the path.

2. Step 2: Identify correct recursive logic

   It should call hasPathSum(root->left, targetSum - root->val) and similarly for right to correctly track remaining sum.

3. Final Answer:

   The targetSum is not reduced by root->val in recursive calls -> Option C

4. Quick Check:

   Subtract node value each step [OK]

Hint: Always subtract current node value from targetSum in recursion [OK]

Common Mistakes:

• Not updating targetSum in recursive calls
• Confusing base case return values
• Returning wrong data type

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?

medium

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

B. Use a global variable to track sums instead of subtracting targetSum

C. Modify the function to return the maximum path sum instead of a boolean

D. Check all paths, not just root-to-leaf, because negative values can cancel out sums

Solution

1. Step 1: Consider negative values in path sum

   The standard recursive algorithm subtracts the current node's value (which can be negative) from the targetSum at each step, correctly handling negatives, zeros, and positives for root-to-leaf paths.

2. Step 2: No change required

   The problem is specifically for root-to-leaf paths, not arbitrary paths, so no modification is needed.

3. Final Answer:

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

4. Quick Check:

   Recursion handles negatives via subtraction [OK]

Hint: Standard path sum recursion works with negative node values [OK]

Common Mistakes:

• Thinking negatives require checking all paths
• Using global variables unnecessarily
• Changing return type without reason