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

Path Sum Root to Leaf in Binary Tree

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

🎯 Goal: You will build a small binary tree with exact ages, set a target sum, write a function to check if any root-to-leaf path sums to the target, and print the result.

📋 What You'll Learn

Create a binary tree with nodes having exact integer values

Set a target sum variable

Write a function called hasPathSum that takes the root node and target sum and returns a boolean

Print the boolean result of hasPathSum

💡 Why This Matters

🌍 Real World

Checking path sums in trees is useful in decision trees, network routing, and financial calculations 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 structure and nodes

Create a struct called TreeNode with fields Val int, Left *TreeNode, and Right *TreeNode. Then create a variable called root representing the root node with value 5. Set its left child to a node with value 4 and its right child to a node with value 8. The left child of the node with value 4 should be a node with value 11. The left and right children of the node with value 8 should be nodes with values 13 and 4 respectively. Finally, the node with value 11 should have left and right children with values 7 and 2. The right child of the node with value 4 (right subtree) should have a right child with value 1.

DSA Go

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

Hint

Start by defining the TreeNode struct. Then create the root node and assign its children exactly as described.

2

Set the target sum variable

Add a variable called targetSum and set it to 22.

DSA Go

package main

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func main() {
    root := &TreeNode{Val: 5}
    root.Left = &TreeNode{Val: 4}
    root.Right = &TreeNode{Val: 8}
    root.Left.Left = &TreeNode{Val: 11}
    root.Left.Left.Left = &TreeNode{Val: 7}
    root.Left.Left.Right = &TreeNode{Val: 2}
    root.Right.Left = &TreeNode{Val: 13}
    root.Right.Right = &TreeNode{Val: 4}
    root.Right.Right.Right = &TreeNode{Val: 1}
    
    // Set the target sum variable
    // Your code here
}

Hint

Just add a new variable targetSum inside the main function and assign it the value 22.

3

Write the hasPathSum function

Write a function called hasPathSum that takes parameters root *TreeNode and sum int and returns a bool. The function should return true if there exists a root-to-leaf path in the tree where the sum of node values equals sum. Otherwise, return false. Use recursion to check the left and right subtrees, subtracting the current node's value from sum. If the node is a leaf (no left and right children), check if the remaining sum equals the node's value.

DSA Go

package main

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func hasPathSum(root *TreeNode, sum int) bool {
    // Your code here
}

func main() {
    root := &TreeNode{Val: 5}
    root.Left = &TreeNode{Val: 4}
    root.Right = &TreeNode{Val: 8}
    root.Left.Left = &TreeNode{Val: 11}
    root.Left.Left.Left = &TreeNode{Val: 7}
    root.Left.Left.Right = &TreeNode{Val: 2}
    root.Right.Left = &TreeNode{Val: 13}
    root.Right.Right = &TreeNode{Val: 4}
    root.Right.Right.Right = &TreeNode{Val: 1}

    targetSum := 22
}

Hint

Use recursion. Check if the node is nil, then if it's a leaf, compare its value to the sum. Otherwise, recursively check left and right children with the updated sum.

4

Print the result of hasPathSum

Add a print statement in the main function to display the result of calling hasPathSum(root, targetSum).

DSA Go

Hint

Use println(hasPathSum(root, targetSum)) inside the main function to show the result.

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 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

   It is not about counting leaves, finding max depth, or summing all nodes, 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 A

4. Quick Check:

   Path sum = target? [OK]

Hint: Focus on root-to-leaf paths, not entire tree sums [OK]

Common Mistakes:

• Confusing path sum with total tree sum
• Checking paths that don't end at leaf nodes
• Ignoring the root node in the sum

2. Which of the following Go function signatures correctly represents a function to check path sum in a binary tree node defined as type TreeNode struct { Val int; Left, Right *TreeNode }?

easy

A. func hasPathSum(root TreeNode, sum int) bool

B. func hasPathSum(root *TreeNode, sum int) bool

C. func hasPathSum(root *TreeNode, sum int) int

D. func hasPathSum(root TreeNode, sum int) int

Solution

1. Step 1: Identify correct parameter types

   The root node should be passed as a pointer to TreeNode to allow nil checks and avoid copying the entire node.

2. Step 2: Identify correct return type

   The function should return a boolean indicating if such a path exists, not an integer.

3. Final Answer:

   func hasPathSum(root *TreeNode, sum int) bool -> Option B

4. Quick Check:

   Pointer param + bool return = correct signature [OK]

Hint: Use pointer receiver and bool return for path existence [OK]

Common Mistakes:

• Passing TreeNode by value instead of pointer
• Returning int instead of bool
• Using incorrect parameter order

3. What is the output of the following Go code snippet?

package main

import "fmt"

type TreeNode struct {
  Val int
  Left, Right *TreeNode
}

func hasPathSum(root *TreeNode, sum int) bool {
  if root == nil {
    return false
  }
  if root.Left == nil && root.Right == nil {
    return root.Val == sum
  }
  return hasPathSum(root.Left, sum-root.Val) || hasPathSum(root.Right, sum-root.Val)
}

func main() {
  root := &TreeNode{5, &TreeNode{4, &TreeNode{11, &TreeNode{7, nil, nil}, &TreeNode{2, nil, nil}}, nil}, &TreeNode{8, &TreeNode{13, nil, nil}, &TreeNode{4, nil, &TreeNode{1, nil, nil}}}}
  fmt.Println(hasPathSum(root, 22))
}

medium

A. Compilation error

B. false

C. Runtime panic

D. true

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, so the function returns true.

2. Step 2: Confirm function logic correctness

   The function correctly checks leaf nodes and recursively subtracts node values from sum, returning true if any path matches.

3. Final Answer:

   true -> Option D

4. Quick Check:

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

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

Common Mistakes:

• Ignoring leaf node condition
• Not subtracting node value from sum
• Assuming sum must be at root only

4. The following Go function is intended to check if a path sum exists but has a bug. What is the bug?

func hasPathSum(root *TreeNode, sum int) bool {
  if root == nil {
    return false
  }
  if root.Left == nil && root.Right == nil {
    return sum == root.Val
  }
  return hasPathSum(root.Left, sum-root.Val) && hasPathSum(root.Right, sum-root.Val)
}

medium

A. Using && instead of || causes incorrect results

B. Missing base case for nil root

C. Incorrect comparison order in leaf check

D. Not subtracting root.Val from sum in recursive calls

Solution

1. Step 1: Analyze recursive return condition

   The function uses logical AND (&&) to combine left and right subtree results, which requires both to be true.

2. Step 2: Understand correct logic for path sum

   We want to return true if either left or right subtree has a valid path, so logical OR (||) should be used instead.

3. Final Answer:

   Using && instead of || causes incorrect results -> Option A

4. Quick Check:

   Use OR (||) to find any valid path [OK]

Hint: Use OR (||) to combine subtree results, not AND (&&) [OK]

Common Mistakes:

• Using AND instead of OR in recursive calls
• Forgetting to check leaf nodes properly
• Not handling nil root correctly

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. Only consider paths with positive sums

B. Stop recursion early if sum becomes negative

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

D. Use breadth-first search instead of depth-first search

Solution

1. Step 1: Understand path sum logic with negative values

   The standard recursive approach subtracts node values from sum and checks leaf nodes; negative values do not break this logic.

2. Step 2: Evaluate unnecessary optimizations

   Stopping early on negative sums or ignoring negative paths can miss valid paths; BFS is not required for correctness.

3. Final Answer:

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

4. Quick Check:

   Standard DFS works with negatives [OK]

Hint: Standard DFS path sum works regardless of node values sign [OK]

Common Mistakes:

• Stopping recursion on negative sums
• Ignoring negative or zero values in paths
• Switching to BFS unnecessarily