Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.
A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
Example 2:
Input: root = [1,2,3], targetSum = 5
Output: []
Example 3:
Input: root = [1,2], targetSum = 0
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 5000]. -1000 <= Node.val <= 1000-1000 <= targetSum <= 1000
/**
* 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 result: [[Int]] = []
traverse(root, targetSum, [], &result)
return result
}
func traverse(_ node: TreeNode?, _ targetSum: Int, _ values: [Int], _ result: inout [[Int]]) {
guard let node else { return }
let targetSum: Int = targetSum - node.val
let values: [Int] = values + [node.val]
if node.left == nil, node.right == nil {
if targetSum == 0 {
result.append(values)
}
} else {
[node.left, node.right].forEach {
traverse($0, targetSum, values, &result)
}
}
}
}