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<>();
        stack.add(root);
        
        while (!stack.empty()) {
            root = stack.pop();
            list.add(root.val);
            for (int i = root.children.size() - 1; i >= 0; i--)
                stack.add(root.children.get(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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