Given an n-ary tree, return the preorder traversal of its nodes' values.
Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).
Follow up: Recursive solution is trivial, could you do it iteratively?
Example 1:
Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]
Example 2:
Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]
Constraints:
- The height of the n-ary tree is less than or equal to
1000
- The total number of nodes is between
[0, 10^4]
O(n) Time, O(1) Space - Recursive:
/**
* Definition for a Node.
* public class Node {
* public var val: Int
* public var children: [Node]
* public init(_ val: Int) {
* self.val = val
* self.children = []
* }
* }
*/
class Solution {
func preorder(_ root: Node?) -> [Int] {
var result : [Int] = []
preOrder(root, &result)
return result
}
func preOrder(_ node: Node?, _ result: inout [Int]) {
guard let node = node else { return }
result.append(node.val)
node.children.forEach {
preOrder($0, &result)
}
}
}
O(n) Time, O(n) Space - Iterative:
/**
* Definition for a Node.
* public class Node {
* public var val: Int
* public var children: [Node]
* public init(_ val: Int) {
* self.val = val
* self.children = []
* }
* }
*/
class Solution {
func preorder(_ root: Node?) -> [Int] {
guard let root = root else { return [] }
var stack = [root], result : [Int] = []
while !stack.isEmpty {
let curr = stack.removeLast()
for i in stride(from: curr.children.count-1, through: 0, by: -1) {
stack.append(curr.children[i])
}
result.append(curr.val)
}
return result
}
}