# leetcode\_589

## 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]

## Solutions

1. **recursion**

```cpp
/*
// 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 {
private:
    vector<int> res;
public:
    void preOrder(Node * root) {
        res.push_back(root->val);
        for (auto & child : root->children)
            if (child)
                preOrder(child);
    }
    vector<int> preorder(Node* root) {
        if (root) preOrder(root);
        return res;
    }
};
```

1. **iteration with stack**

```cpp
/*
// 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> preorder(Node* root) {
        vector<int> res;
        stack<Node *> s;
        if (root) s.push(root);

        while (!s.empty()) {
            root = s.top(); s.pop();
            res.push_back(root->val);
            auto rit = root->children.rbegin();
            while (rit != root->children.rend())
                if (*rit) s.push(*rit++);
        }

        return res;
    }
};
```