LeetCode in Kotlin
2415. Reverse Odd Levels of Binary Tree
Medium
Given the root of a perfect binary tree, reverse the node values at each odd level of the tree.
- For example, suppose the node values at level 3 are
[2,1,3,4,7,11,29,18], then it should become[18,29,11,7,4,3,1,2].
Return the root of the reversed tree.
A binary tree is perfect if all parent nodes have two children and all leaves are on the same level.
The level of a node is the number of edges along the path between it and the root node.
Example 1:
Input: root = [2,3,5,8,13,21,34]
Output: [2,5,3,8,13,21,34]
Explanation:
The tree has only one odd level.
The nodes at level 1 are 3, 5 respectively, which are reversed and become 5, 3.
Example 2:
Input: root = [7,13,11]
Output: [7,11,13]
Explanation:
The nodes at level 1 are 13, 11, which are reversed and become 11, 13.
Example 3:
Input: root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
Output: [0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
Explanation:
The odd levels have non-zero values.
The nodes at level 1 were 1, 2, and are 2, 1 after the reversal.
The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1 after the reversal.
Constraints:
- The number of nodes in the tree is in the range
[1, 214]. 0 <= Node.val <= 105rootis a perfect binary tree.
Solution
import com_github_leetcode.TreeNode
import java.util.LinkedList
import java.util.Queue
/*
* 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 list: MutableList<Int> = ArrayList()
fun reverseOddLevels(root: TreeNode): TreeNode? {
solve(root)
return enrich(list, 0)
}
private fun enrich(list: List<Int>, i: Int): TreeNode? {
var root: TreeNode? = null
if (i < list.size) {
root = TreeNode(list[i])
root.left = enrich(list, 2 * i + 1)
root.right = enrich(list, 2 * i + 2)
}
return root
}
private fun solve(root: TreeNode) {
val q: Queue<TreeNode?> = LinkedList()
q.add(root)
var level = 0
while (q.isNotEmpty()) {
val size = q.size
val res: MutableList<Int> = ArrayList()
for (i in 0 until size) {
val cur = q.remove()
res.add(cur!!.`val`)
if (cur.left != null) {
q.add(cur.left)
}
if (cur.right != null) {
q.add(cur.right)
}
}
if (level % 2 != 0) {
for (i in res.indices.reversed()) {
list.add(res[i])
}
} else {
list.addAll(res)
}
level++
}
}
}