LeetCode-in-Swift
437. Path Sum III
Medium
Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.
The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
Example 1:
Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
Output: 3
Explanation: The paths that sum to 8 are shown.
Example 2:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: 3
Constraints:
- The number of nodes in the tree is in the range
[0, 1000]. -109 <= Node.val <= 109-1000 <= targetSum <= 1000
Solution
/**
* Definition for a binary tree node.
* public class TreeNode {
* public var val: Int
* public var left: TreeNode?
* public var right: TreeNode?
* public init() { self.val = 0; self.left = nil; self.right = nil; }
* public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
* public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
* self.val = val
* self.left = left
* self.right = right
* }
* }
*/
class Solution {
func pathSum(_ root: TreeNode?, _ targetSum: Int) -> Int {
var count = 0
var prefixSums: [Int: Int] = [:]
var prefixSum = 0
func preOrderTraversal(_ root: TreeNode?, _ prefixSum: inout Int, _ prefixSums: inout [Int: Int]) {
guard let root = root else { return }
prefixSum += root.val
count += prefixSum == targetSum ? 1 : 0
count += prefixSums[prefixSum - targetSum] ?? 0
prefixSums[prefixSum, default: 0] += 1
preOrderTraversal(root.left, &prefixSum, &prefixSums)
preOrderTraversal(root.right, &prefixSum, &prefixSums)
prefixSums[prefixSum, default: 0] -= 1
prefixSum -= root.val
}
preOrderTraversal(root, &prefixSum, &prefixSums)
return count
}
}