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# N-ary Tree Preorder Traversal LeetCode Solution

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

Given the `root` of 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)

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 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 Preorder Traversal LeetCode Solution in Java

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

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

while (!stack.empty()) {
root = stack.pop();
for (int i = root.children.size() - 1; i >= 0; i--)
}

return list;
}
}
``````

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

``````class Solution {
private:
void travel(Node* root, vector<int>& result) {
if (root == nullptr) {
return;
}

result.push_back(root -> val);
for (Node* child : root -> children) {
travel(child, result);
}
}
public:
vector<int> preorder(Node* root) {
vector<int> result;
travel(root, result);
return result;
}
};
``````

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

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

class Solution(object):
def preorder(self, root):
"""
:type root: Node
:rtype: List[int]
"""

output =[]

# perform dfs on the root and get the output stack
self.dfs(root, output)

# return the output of all the nodes.
return output

def dfs(self, root, output):

# If root is none return
if root is None:
return

# for preorder we first add the root val
output.append(root.val)

# Then add all the children to the output
for child in root.children:
self.dfs(child, output)

``````
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