LeetCode in Kotlin
103. Binary Tree Zigzag Level Order Traversal
Medium
Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]
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]. -100 <= Node.val <= 100
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 {
fun zigzagLevelOrder(root: TreeNode?): List<List<Int>> {
val result: MutableList<List<Int>> = ArrayList()
if (root == null) {
return result
}
val q: Queue<TreeNode> = LinkedList()
q.add(root)
q.add(null)
var zig = true
var level = LinkedList<Int>()
while (q.isNotEmpty()) {
var node: TreeNode? = q.remove()
while (q.isNotEmpty() && node != null) {
if (zig) {
level.add(node.`val`)
} else {
level.addFirst(node.`val`)
}
if (node.left != null) {
q.add(node.left)
}
if (node.right != null) {
q.add(node.right)
}
node = q.remove()
}
result.add(level)
zig = !zig
level = LinkedList()
if (q.isNotEmpty()) {
q.add(null)
}
}
return result
}
}