Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example, given a 3-ary tree:

We should return its level order traversal:

[
[1],
[3,2,4],
[5,6]
]

Note:

  1. The depth of the tree is at most 1000.
  2. The total number of nodes is at most 5000.

-----------------------------------------------------------------------------------------------------------------------------------

这个层序遍历,自然用BFS写。和二叉树的层序遍历类似,连代码不会相差很大。

C++代码:

/*
// 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) {
if(!root) return {};
queue<Node*> q;
q.push(root);
vector<vector<int> > vec;
while(!q.empty()){
vector<int> vec1;
int ans = q.size();
for(int i = ans;i > ; i--){
auto t = q.front();
q.pop();
vec1.push_back(t->val);
for(Node *cur:t->children){
q.push(cur);
}
}
vec.push_back(vec1);
}
return vec;
}
};

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