LeetCode in Kotlin
113. Path Sum II
Medium
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
Solution
import com_github_leetcode.TreeNode
/*
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
@Suppress("NAME_SHADOWING")
class Solution {
fun pathSum(root: TreeNode?, targetSum: Int): List<List<Int>> {
val res: MutableList<List<Int>> = ArrayList()
if (root == null) {
return res
}
recur(res, ArrayList(), 0, targetSum, root)
return res
}
private fun recur(
res: MutableList<List<Int>>,
al: ArrayList<Int>,
sum: Int,
targetSum: Int,
root: TreeNode?,
) {
var sum = sum
if (root == null) {
return
}
al.add(root.`val`)
sum += root.`val`
if (sum == targetSum && root.left == null && root.right == null) {
res.add(ArrayList(al))
}
recur(res, al, sum, targetSum, root.left)
recur(res, al, sum, targetSum, root.right)
al.removeAt(al.size - 1)
}
}