590 N-ary Tree Postorder Traversal N叉树的后序遍历

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

Example:

For example, given a 3-ary tree:

3-ary tree

Return its postorder traversal as: [5,6,3,2,4,1].

Note:

Recursive solution is trivial, could you do it iteratively?

题目描述:
给定一个 N 叉树，返回其节点值的后序遍历。

示例 :

例如，给定一个 3叉树 :

3叉树

返回其后序遍历: [5,6,3,2,4,1]。

说明:
递归法很简单，你可以使用迭代法完成此题吗?

思路:

1. 递归
2. 迭代, 利用栈, 插入结点的孩子结点

时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution 
{
public:
    vector<int> postorder(Node* root) 
    {
        vector<int> result;
        if (!root) return result;
        stack<Node*> s;
        s.push(root);
        while (s.size()) 
        {
            auto cur = s.top();
            s.pop();
            result.insert(result.begin(), cur -> val);
            for (auto temp : cur -> children) s.push(temp);
        }
        return result;
    }
};

Java:

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val,List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
class Solution {
    public List<Integer> postorder(Node root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) return result;
        for (Node node : root.children) result.addAll(postorder(node));
        result.add(root.val);
        return result;
    }
}

Python:

"""
# Definition for a Node.
class Node:
    def __init__(self, val, children):
        self.val = val
        self.children = children
"""
class Solution:
    def postorder(self, root: 'Node') -> List[int]:
        result = []
        def helper(root: 'Node') -> None:
            if not root:
                return
            for item in root.children:
                helper(item)
            result.append(root.val)
        helper(root)
        return result