From 2c830c0cd129232e3a7325000e24d8c597e0f39c Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Wed, 14 Nov 2018 23:53:14 +0800
Subject: [PATCH] Create 429. N-ary Tree Level Order Traversal.md

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+# [429. N-ary Tree Level Order Traversal](https://leetcode.com/problems/n-ary-tree-level-order-traversal/description/)
+# 思路
+树的层次遍历。和二叉树的层次遍历其实是一样的。    
+用last指针指向每一层的最后一个节点，每当遍历到这个节点即说明遍历完一层，
+此时应该将此层所有节点（用数组a_level记录）push进保存最终结果的数组res里，然后清空a_level，继续遍历下一层。
+last初始为root， 后面每当遍历完每层最后一个节点后，即将last更新成下一层的最后一个节点，为此需要用一个next_last来不断记录能确定的下一层的最右节点。   
+时间复杂度和空间复杂度都是O(n)  
+(讨论区有一个[运行时间比较短的递归算法](https://leetcode.com/problems/n-ary-tree-level-order-traversal/discuss/157521/C++-Easy-to-understand-recursive-solution-based-on-DFS-(44-ms-beats-98.67))，但是评论说其复杂度比较高，所以没细看)
+
+# C++
+``` C++
+class Solution {
+public:
+    vector<vector<int>> levelOrder(Node* root) {
+        vector<vector<int>>res;
+        if(!root) return res;
+        
+        queue<Node*>q;
+        vector<int>a_level;
+        Node *p, *next_last, *last = root;
+        q.push(root);
+        while(!q.empty()){
+            p = q.front();
+            q.pop();
+            
+            a_level.push_back(p -> val);
+            
+            for(int i = 0; i < p -> children.size(); i++){
+                next_last = p -> children[i];
+                q.push(next_last);
+            }
+            
+            if(p == last){
+                res.push_back(a_level);
+                a_level.clear();
+                last = next_last;
+            }
+        }
+        return res;        
+    }
+};
+```