Path Sum Root to Leaf in Binary Tree in DSA Go - Time & Space Complexity

Time Complexity: Path Sum Root to Leaf in Binary Tree

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a path from the root to any leaf adds up to a target sum in a binary tree.

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code checks if there is any root-to-leaf path where the sum of node values equals the target sum.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree, covering all paths.

How Execution Grows With Input

As the number of nodes grows, the function visits each node once, so the work grows proportionally.

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

Pattern observation: The operations grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check the path sum grows directly 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 root-to-leaf paths by visiting every node, so it must check many nodes, not just one path.

Interview Connect

Understanding how recursive tree traversal scales helps you explain your solution clearly and shows you grasp how algorithms work on trees.

Self-Check

"What if the tree was balanced versus completely skewed? How would the time complexity change?"

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