树590. N叉树的后序遍历

Posted 2021-02-20 ocpc

题目:

 

 

解答：

方法一：递归

 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     vector<int> postorder(Node* root) 
24     {
25         vector<int> result;
26         if (NULL == root)
27         {
28             return result;
29         }
30 
31         recursivePostorder(root, result);
32         return result;
33     }
34 
35     void recursivePostorder(Node *root, vector<int> &ret)
36     {
37         if (NULL == root)
38         {
39             return;
40         }
41 
42         int size = root->children.size();
43         for (int i = 0; i < size; i++)
44         {
45             Node *n = root->children[i];
46             if (n)
47             {
48                 recursivePostorder(n, ret);
49             }
50         }
51 
52         ret.push_back(root->val);
53     } 
54 };

 

方法二：迭代

 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      // 第二种：迭代, 先前序，然后反转
24    vector<int> postorder(Node* root) {
25         vector<int> ve;
26         if (!root) return ve;
27 
28         stack<Node*> st;
29         st.push(root);
30         
31         while (!st.empty()) {
32             Node *node = st.top();
33             st.pop();
34 
35             if (node) {
36                 ve.emplace_back(node->val);
37                 
38                 vector<Node*> chs = node->children;
39                 if (!chs.empty()) {
40                     int size = chs.size();
41                     for (int i =0; i< size; i++) {
42                         Node *n =  chs[i];
43                         if (n) st.push(n);
44                     }
45                 }
46             }
47         }
48     
49         reverse(ve.begin(),ve.end());
50         return ve;
51     }
52 
53 };