# Daily Coding Challenge #75

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

/*
LeetCode - N-ary Tree Preorder Traversal
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).

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]
*/

/*
// 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:
void solve(Node* root, vector<int>& ans){
if(!root) return;
// pre order
ans.emplace_back(root->val);
for(auto c:root->children){
solve(c,ans);
}
}

vector<int> preorder(Node* root) {
vector<int> ans;
solve(root,ans);
return ans;
}
};


/*
LeetCode - N-ary Tree Postorder Traversal

Given an n-ary tree, return the postorder 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).

Recursive solution is trivial, could you do it iteratively?

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [5,6,3,2,4,1]

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: [2,6,14,11,7,3,12,8,4,13,9,10,5,1]

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]
*/

/*
// 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:
void solve(Node* root, vector<int>& ans){
if(!root) return;
for(auto c:root->children) solve(c, ans);
// post order
ans.emplace_back(root->val);
}
vector<int> postorder(Node* root) {
vector<int> ans;
solve(root,ans);
return ans;
}
};


/*
LeetCode - N-ary Tree Level Order Traversal

Given an n-ary tree, return the level order 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],[3,2,4],[5,6]]

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],[2,3,4,5],[6,7,8,9,10],[11,12,13],[14]]

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]
*/

/*
// 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<vector<int>> levelOrder(Node* root) {
// similar to 102 & 103
// bfs approach
vector<vector<int>> ans;
if(!root) return ans;
queue<Node*> q;
q.push(root);
while(!q.empty()){
int n = q.size();
vector<int> cur(n);
for(int i=0;i<n;i++){
Node* tmp = q.front();
q.pop();
cur[i] = tmp->val;
vector<Node*> children = tmp->children;
for(auto c:children) q.push(c);
}
ans.emplace_back(cur);
}
return ans;
}
};


The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

Posted on by:

### Wing-Kam

Consultant by day. Developer by night. AWS certified. Exploring #CloudNative currently.