题目描述
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).
Example1:
Input: root = [1,null,3,2,4,null,5,6] Output: [[1],[3,2,4],[5,6]]
Example2:
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]]
解题思路
- 类似于BFS遍历多叉树
/*
// 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) {
if(root == nullptr){
return {};
}
vector<vector<int>> res;
queue<Node*> q;
q.push(root);
while(!q.empty()){
int size = q.size();
vector<int> temp;
while(size != 0){
Node* node = q.front();
temp.push_back(node -> val);
for(int i = 0; i < node -> children.size(); i++){
q.push(node -> children[i]);
}
q.pop();
size--;
}
res.push_back(temp);
}
return res;
}
};
