590. N 叉树的后序遍历

n 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 null 分隔（请参见示例）。

示例 1：

输入：root = [1,null,3,2,4,null,5,6]
输出：[5,6,3,2,4,1]

示例 2：

输入：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]
输出：[2,6,14,11,7,3,12,8,4,13,9,10,5,1]

提示：

• 节点总数在范围 [0, 104] 内
• 0 <= Node.val <= 104
• n 叉树的高度小于或等于 1000

递归解法：

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4 public:
 5     int val;
 6     vector<Node*> children;
 7 
 8     Node() {}
 9 
10     Node(int _val) {
11         val = _val;
12     }
13 
14     Node(int _val, vector<Node*> _children) {
15         val = _val;
16         children = _children;
17     }
18 };
19 */
20 
21 class Solution {
22 public:
23     void postorderHelper(Node *root, vector<int> &vec) {
24         if (root == nullptr) {
25             return;
26         }
27         int size = root->children.size();
28         for (int i = 0; i < size; i++) {
29             postorderHelper(root->children[i], vec);           
30         }
31         vec.push_back(root->val);
32         return;
33     }
34     vector<int> postorder(Node* root) {
35         vector<int> ans;
36         if (root == nullptr) {
37             return ans;
38         }        
39         postorderHelper(root, ans);
40         return ans;
41     }
42 };

迭代解法：

 1 class Solution {
 2 public:
 3     vector<int> postorder(Node* root) {
 4         vector<int> ans;
 5         if (root == nullptr) {
 6             return ans;
 7         }        
 8         stack<Node*> s;
 9         s.push(root);
10         while (!s.empty()) {
11             Node * p = s.top();
12             s.pop();
13             if (p != nullptr) {
14                 s.push(p);
15                 s.push(nullptr);
16                 int size = p->children.size();
17                 for (int i = size - 1; i >= 0; i--) {
18                     s.push(p->children[i]);
19                 } 
20             } else {
21                 p = s.top();
22                 s.pop();
23                 ans.push_back(p->val);
24             }
25         }
26         return ans;
27     }
28 };