What if you could instantly know if a path adds up to your target without checking every step yourself?
Why Path Sum Root to Leaf in Binary Tree in DSA Go?
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 hand means checking every possible path, adding numbers, and hoping you don't miss any.
Manually adding numbers along every path is slow and easy to mess up, especially if the tree is big.
You might forget a path or add wrong numbers, making the whole process frustrating and error-prone.
Using a program to check the tree automatically saves time and avoids mistakes.
The program walks down each path, keeps track of the sum, and tells you if any path matches the target sum.
func checkPathSum(root *Node, target int) bool {
// Manually check each path by hand
return false // too complex to do manually
}func hasPathSum(root *Node, 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 lets you quickly find if any root-to-leaf path adds up to a target, even in huge trees.
Checking if a route in a map adds up to a certain distance or cost, helping in navigation or planning.
Manual checking is slow and error-prone.
Recursive path sum checking automates and simplifies the task.
It helps find exact paths matching a sum quickly.
Practice
Solution
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.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.Final Answer:
If there exists a root-to-leaf path with a sum equal to a given target -> Option AQuick Check:
Path sum = target? [OK]
- Confusing path sum with total tree sum
- Checking paths that don't end at leaf nodes
- Ignoring the root node in the sum
type TreeNode struct { Val int; Left, Right *TreeNode }?Solution
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.Step 2: Identify correct return type
The function should return a boolean indicating if such a path exists, not an integer.Final Answer:
func hasPathSum(root *TreeNode, sum int) bool -> Option BQuick Check:
Pointer param + bool return = correct signature [OK]
- Passing TreeNode by value instead of pointer
- Returning int instead of bool
- Using incorrect parameter order
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))
}Solution
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.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.Final Answer:
true -> Option DQuick Check:
Path 5+4+11+2 = 22 [OK]
- Ignoring leaf node condition
- Not subtracting node value from sum
- Assuming sum must be at root only
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)
}Solution
Step 1: Analyze recursive return condition
The function uses logical AND (&&) to combine left and right subtree results, which requires both to be true.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.Final Answer:
Using && instead of || causes incorrect results -> Option AQuick Check:
Use OR (||) to find any valid path [OK]
- Using AND instead of OR in recursive calls
- Forgetting to check leaf nodes properly
- Not handling nil root correctly
Solution
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.Step 2: Evaluate unnecessary optimizations
Stopping early on negative sums or ignoring negative paths can miss valid paths; BFS is not required for correctness.Final Answer:
No modification needed; the standard path sum check works for all values -> Option CQuick Check:
Standard DFS works with negatives [OK]
- Stopping recursion on negative sums
- Ignoring negative or zero values in paths
- Switching to BFS unnecessarily
Ready to prove it?
Pass the test and this pattern moves unknown
