LeetCode in Kotlin
107. Binary Tree Level Order Traversal II
Medium
Given the root of a binary tree, return the bottom-up level order traversal of its nodes’ values. (i.e., from left to right, level by level from leaf to root).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],[3]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -1000 <= Node.val <= 1000
Solution
import com_github_leetcode.TreeNode
import java.util.Collections
import kotlin.collections.ArrayList
/*
* 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 {
private val order: MutableList<MutableList<Int>> = ArrayList()
fun levelOrderBottom(root: TreeNode?): List<MutableList<Int>> {
getOrder(root, 0)
Collections.reverse(order)
return order
}
fun getOrder(root: TreeNode?, level: Int) {
if (root != null) {
if (level + 1 > order.size) {
order.add(ArrayList())
}
order[level].add(root.`val`)
getOrder(root.left, level + 1)
getOrder(root.right, level + 1)
}
}
}