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# N-ary Tree Postorder Traversal LeetCode Solution

## Problem – N-ary Tree Postorder Traversal LeetCode Solution

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`.

### N-ary Tree Postorder Traversal LeetCode Solution in Java

``````class Solution {
public List<Integer> postorder(Node root) {
List<Integer> list = new ArrayList<>();
if (root == null) return list;

Stack<Node> stack = new Stack<>();

while(!stack.isEmpty()) {
root = stack.pop();
for(Node node: root.children)
}
Collections.reverse(list);
return list;
}
}
``````

### N-ary Tree Postorder Traversal LeetCode Solution in C++

``````vector<int> postorder(Node* root) {
if(root==NULL) return {};
vector<int> res;
stack<Node*> stk;
stk.push(root);
while(!stk.empty())
{
Node* temp=stk.top();
stk.pop();
for(int i=0;i<temp->children.size();i++) stk.push(temp->children[i]);
res.push_back(temp->val);
}
reverse(res.begin(), res.end());
return res;
}
``````

### N-ary Tree Postorder Traversal LeetCode Solution in Python

``````def postorder(self, root):
"""
:type root: Node
:rtype: List[int]
"""
res = []
if root == None: return res

def recursion(root, res):
for child in root.children:
recursion(child, res)
res.append(root.val)

recursion(root, res)
return res
``````
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