LeetCode 590. N-ary Tree Postorder Traversal

Given the root of an n-ary tree, return the postorder traversal of its nodes' values.

Nary-Tree input serialization is represented in their level order traversal. Each group of children is separated by the null value (See examples)

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [5,6,3,2,4,1]

Example 2:

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

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• 0 <= Node.val <= 104
• The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?

递归，也是很简单，只是变成了先遍历children再加root.val。

1.  
   /*
2.  
   // Definition for a Node.
3.  
   class Node {
4.  
   public int val;
5.  
   public List<Node> children;
6.  
    
7.  
   public Node() {}
8.  
    
9.  
   public Node(int _val) {
10.  
    val = _val;
11.  
    }
12.  
     
13.  
    public Node(int _val, List<Node> _children) {
14.  
    val = _val;
15.  
    children = _children;
16.  
    }
17.  
    };
18.  
    */
19.  
     
20.  
    class Solution {
21.  
    public List<Integer> postorder(Node root) {
22.  
    List<Integer> result = new ArrayList<>();
23.  
    helper(root, result);
24.  
    return result;
25.  
    }
26.  
     
27.  
    public void helper(Node root, List<Integer> result) {
28.  
    if (root == null) {
29.  
    return;
30.  
    }
31.  
    for (Node node : root.children) {
32.  
    helper(node, result);
33.  
    }
34.  
    result.add(root.val);
35.  
    }
36.  
    }

迭代，还是记着Stack Set的方法呢，但是，debug了巨久都没搞出来，大概是seen set的地方弄错了。想着应该要把所有children都seen了才能pop它，但是刚开始想的是seen表示这个node有没有见过，于是要check所有children都seen了还挺麻烦，也不太确定这个思路对不对，就去看了答案。居然几乎没什么人跟我一个思路，看了一个相似的代码，才确认了确实是只有当node没有chidlren或者children全都seen过了才能pop，于是就把seen的含义变成了这个node的所有children是不是都seen过，调整了seen.add的位置，代码就很简洁也一下就没bug了。也算是加深了对postorder iterative的理解。

1.  
   /*
2.  
   // Definition for a Node.
3.  
   class Node {
4.  
   public int val;
5.  
   public List<Node> children;
6.  
    
7.  
   public Node() {}
8.  
    
9.  
   public Node(int _val) {
10.  
    val = _val;
11.  
    }
12.  
     
13.  
    public Node(int _val, List<Node> _children) {
14.  
    val = _val;
15.  
    children = _children;
16.  
    }
17.  
    };
18.  
    */
19.  
     
20.  
    class Solution {
21.  
    public List<Integer> postorder(Node root) {
22.  
    List<Integer> result = new ArrayList<>();
23.  
    if (root == null) {
24.  
    return result;
25.  
    }
26.  
    Deque<Node> stack = new ArrayDeque<>();
27.  
    Set<Node> seen = new HashSet<>(); // this node's children are all seen
28.  
    stack.push(root);
29.  
    while (!stack.isEmpty()) {
30.  
    Node node = stack.peek();
31.  
    int size = node.children.size();
32.  
    if (size == 0 || seen.contains(node)) {
33.  
    node = stack.pop();
34.  
    result.add(node.val);
35.  
    } else {
36.  
    for (int i = size - 1; i >= 0; i--) {
37.  
    Node child = node.children.get(i);
38.  
    if (!seen.contains(child)) {
39.  
    stack.push(child);
40.  
    }
41.  
    }
42.  
    seen.add(node);
43.  
    }
44.  
    }
45.  
    return result;
46.  
    }
47.  
    }

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