LeetCode in Kotlin
144. Binary Tree Preorder Traversal
Easy
Given the root of a binary tree, return the preorder traversal of its nodes’ values.
Example 1:
Input: root = [1,null,2,3]
Output: [1,2,3]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
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 preorderTraversal(root: TreeNode?): List<Int> {
val result: MutableList<Int> = ArrayList()
if (root == null) {
return result
}
val stack: ArrayDeque<TreeNode?> = ArrayDeque()
var current: TreeNode? = root
while (current != null || stack.isNotEmpty()) {
while (current != null) {
result.add(current.`val`)
stack.addLast(current.right)
current = current.left
}
current = stack.removeLast()
}
return result
}
}