LeetCode in Kotlin
1080. Insufficient Nodes in Root to Leaf Paths
Medium
Given the root of a binary tree and an integer limit, delete all insufficient nodes in the tree simultaneously, and return the root of the resulting binary tree.
A node is insufficient if every root to leaf path intersecting this node has a sum strictly less than limit.
A leaf is a node with no children.
Example 1:
Input: root = [1,2,3,4,-99,-99,7,8,9,-99,-99,12,13,-99,14], limit = 1
Output: [1,2,3,4,null,null,7,8,9,null,14]
Example 2:
Input: root = [5,4,8,11,null,17,4,7,1,null,null,5,3], limit = 22
Output: [5,4,8,11,null,17,4,7,null,null,null,5]
Example 3:
Input: root = [1,2,-3,-5,null,4,null], limit = -1
Output: [1,null,-3,4]
Constraints:
- The number of nodes in the tree is in the range
[1, 5000]. -105 <= Node.val <= 105-109 <= limit <= 109
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
* }
*/
class Solution {
fun sufficientSubset(root: TreeNode?, limit: Int): TreeNode? {
return if (sufficientSubset(root, limit, 0, root!!.left == null && root.right == null) < limit) null else root
}
fun sufficientSubset(root: TreeNode?, limit: Int, sum: Int, isLeaf: Boolean): Int {
if (root != null) {
val leftSum = sufficientSubset(
root.left,
limit,
sum + root.`val`,
root.left == null && root.right == null
)
val rightSum = sufficientSubset(
root.right,
limit,
sum + root.`val`,
root.left == null && root.right == null
)
if (leftSum < limit) {
root.left = null
}
if (rightSum < limit) {
root.right = null
}
return leftSum.coerceAtLeast(rightSum)
}
return if (isLeaf) sum else Int.MIN_VALUE
}
}