# 590. N-ary Tree Postorder Traversal

## Description

Given the root of an n-ary tree, return the postorder 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)

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [5,6,3,2,4,1]


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: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]


Constraints:

• The number of nodes in the tree is in the range [0, 104].
• 0 <= Node.val <= 104
• The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?

## Solutions

• /*
// Definition for a Node.
class Node {
public int val;
public List<Node> children;

public Node() {}

public Node(int _val) {
val = _val;
}

public Node(int _val, List<Node> _children) {
val = _val;
children = _children;
}
};
*/

class Solution {
public List<Integer> postorder(Node root) {
if (root == null) {
return ans;
}
Deque<Node> stk = new ArrayDeque<>();
stk.offer(root);
while (!stk.isEmpty()) {
root = stk.pollLast();
for (Node child : root.children) {
stk.offer(child);
}
}
return ans;
}
}

• /*
// Definition for a Node.
class Node {
public:
int val;
vector<Node*> children;

Node() {}

Node(int _val) {
val = _val;
}

Node(int _val, vector<Node*> _children) {
val = _val;
children = _children;
}
};
*/

class Solution {
public:
vector<int> postorder(Node* root) {
vector<int> ans;
if (!root) return ans;
stack<Node*> stk{ {root} };
while (!stk.empty()) {
root = stk.top();
ans.push_back(root->val);
stk.pop();
for (Node* child : root->children) stk.push(child);
}
reverse(ans.begin(), ans.end());
return ans;
}
};

• """
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""

class Solution:
def postorder(self, root: 'Node') -> List[int]:
ans = []
if root is None:
return ans
stk = [root]
while stk:
node = stk.pop()
ans.append(node.val)
for child in node.children:
stk.append(child)
return ans[::-1]


• /**
* Definition for a Node.
* type Node struct {
*     Val int
*     Children []*Node
* }
*/

func postorder(root *Node) []int {
var ans []int
if root == nil {
return ans
}
stk := []*Node{root}
for len(stk) > 0 {
root = stk[len(stk)-1]
stk = stk[:len(stk)-1]
ans = append([]int{root.Val}, ans...)
for _, child := range root.Children {
stk = append(stk, child)
}
}
return ans
}

• /**
* Definition for node.
* class Node {
*     val: number
*     children: Node[]
*     constructor(val?: number) {
*         this.val = (val===undefined ? 0 : val)
*         this.children = []
*     }
* }
*/

function postorder(root: Node | null): number[] {
const res = [];
const dfs = (root: Node | null) => {
if (root == null) {
return;
}
for (const node of root.children) {
dfs(node);
}
res.push(root.val);
};
dfs(root);
return res;
}