0500 - 0599
0589 - N-ary Tree Preorder Traversal (Easy)
Problem Link
https://leetcode.com/problems/n-ary-tree-preorder-traversal/
Problem Statement
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, 10^4]. 0 <= Node.val <= 10^4- The height of the n-ary tree is less than or equal to
1000.
Approach 1: DFS
Straightforward preorder traversal.
class Solution {
public:
vector<int> ans;
void dfs(Node* node) {
if (!node) return;
ans.push_back(node->val);
for (auto x : node->children) dfs(x);
}
vector<int> preorder(Node* root) {
dfs(root);
return ans;
}
};"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def preorder(self, root: 'Node') -> List[int]:
output = []
self.dfs(root, output)
return output
def dfs(self, root, output):
if not root:
return
output.append(root.val)
for child in root.children:
self.dfs(child, output)/**
* // Definition for a Node.
* function Node(val, children) {
* this.val = val;
* this.children = children;
* };
*/
/**
* @param {Node|null} root
* @return {number[]}
*/
var preorder = function (root) {
let output = new Array();
dfs(root, output);
return output;
function dfs(root, output) {
if (!root) return;
output.push(root.val);
for (child of root.children) {
dfs(child, output);
}
}
};