刷代码随想录有感（38）：N叉树的层序遍历

题干：

代码：

// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;
    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};


class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        queue<Node*>que;
        vector<vector<int>>res;
        if(root != NULL)que.push(root);
        while(!que.empty()){
            int size = que.size();
            vector<int>tmp;
            while(size--){
                Node* node = que.front();
                que.pop();
                tmp.push_back(node -> val);
                for(int i = 0; i < node -> children.size(); i++){//node->children.size()指当前节点的孩子数目
                    if(node -> children[i]) que.push(node -> children[i]);
                }
            }
            res.push_back(tmp);
        }
        return res;
    }
};

与层序遍历模板不同在于，将if(node->left(right))que.push(node->left(right))改为了：

for(int i = 0; i < node -> children.size(); i++){
    if(node -> children[i]) que.push(node -> children[i]);
} 

node->children.size()指的是当前节点的所有孩子节点个数

if(node -> children[i])指的是如果能扫到节点（即节点存在）就执行后面的步骤。