LeetCode #429 N-aryTreeLevelOrderTraversal. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
문제풀이 설계가 헷갈려서 좀 걸리고 코딩은 금방 끝냄, input 이 null 인 케이스에서 런타임 에러나는걸 다른건줄 알고 삽질 잠깐함
성능은 중간정도 가고, 메모리는 좋지않음. 트리구조랑 재귀 간만에 해서 재밌었음
Spent a lil time to get how to solve, but did coding in a short time. A case input null, i was confused about that runtime error, so spent some time.
Performance is just middle, memory is not that good enough. Anyway, it was so fun to solve cuz it's been a long time to see recursive and tree.
LeetCode #429
Q.
Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
주어진 n 개짜리 트리구조에서, 각 층의 노드 값들을 반환해라. (왼쪽부터 오른쪽으로 층 순서대로)
e.g.
Example:
given a 3-ary
tree:
We should return its level order traversal:
[
[1],
[3,2,4],
[5,6]
]
Process
// Process
//1. Input root node
//2. Call recursiveFunction
//3. Return resultVector
// recursiveFunction
//1. Input node, resultVector, level
//2. Add level
//3. Check if resultVector has that level vector
// 3.1. If not -> make new vector and pushBack
// 3.2. If so -> add node val
//4. Iterate children of node from begin to the end
// 4.1. Call recursive
//5. Subtract level
// 처리과정
//1. 루트노드를 입력받는다.
//2. 재귀함수를 호출한다.
//3. 결과 배열을 반환한다.
// 재귀함수
//1. 노드, 결과배열, 층(레벨) 을 입력받는다.
//2. 층을 더한다.
//3. 결과배열에 해당 층 벡터가 있는지 확인해서
// 3.1. 없으면 -> 만들어서 추가해준다.
// 3.2. 있으면 -> 값을 넣어준다.
//4. 노드의 하위노드에서 전체 반복한다.
// 4.1. 재귀함수 호출한다.
//5. 층을 빼서 돌아간다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
#include <vector>
#include <iostream>
using namespace std;
// Definition for a Node.
class Node {
public:
int val;
vector<Node*> children;
Node() {}
Node(int _val, vector<Node*> _children) {
val = _val;
children = _children;
}
};
class Solution {
public:
vector<vector<int>> levelOrder(Node* root) {
vector<vector<int>> resultVector;
int level = 0;
if (root != 0)
recursiveLevelOrderTraversal(root, &resultVector, &level);
return resultVector;
}
private:
void recursiveLevelOrderTraversal(Node* node, vector<vector<int>>* resultVector, int* level) {
++*level;
if (resultVector->size() < *level)
{
vector<int> *newVector = new vector<int>();
newVector->push_back(node->val);
resultVector->push_back(*newVector);
}
else
{
resultVector->at(*level - 1).push_back(node->val);
}
for (int i = 0; i < node->children.size(); ++i)
{
recursiveLevelOrderTraversal(node->children.at(i), resultVector, level);
}
--*level;
}
};
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