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### Problem:

Given the root of an n-ary tree, return the preorder traversal of its nodes' values.

N ary-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 | [1,3,5,6,2,4] |

Visual |

Example 2 | |
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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] |

Visual |

###### Constraints:

The number of nodes in the tree is in the range

[0, 10^4].

0 <= Node.val <= 10^4

The height of the n-ary tree is less than or equal to1000.

### Idea/Approach:

Pre-order traversal is one type of Depth-First Search (DFS) algorithm. We can solve this problem by using the recursive DFS option. ( add the root(parent) to the result and call DFS recursively for each child)

### C++ Solution:

```
class Solution {
public:
vector<int> result;
void preorderutil(Node * root) {
result.push_back(root->val);
int size = root->children.size();
for(int i = 0; i< size; i++)
preorderutil(root->children[i]);
}
vector<int> preorder(Node* root) {
if(!root)
return {};
preorderutil(root);
return result;
}
};
```

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