589. N-ary Tree Preorder Traversal & 590. N-ary Tree Postorder Traversal

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

For example, given a 3-ary tree:

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

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

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

For example, given a 3-ary tree:

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

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

循环写法不太好用之前二叉树的方法写：https://blog.csdn.net/zjucor/article/details/53447464

因为一直往左边遍历完自后，我是知道要往右边的，而且直接root.right就访问到了，但是在这里是不行的，遍历完第1个child Node，是无法直接得到下个要遍历的节点。

所以要一次性就把children加到stack的解法

class Solution(object):
    def preorder(self, root):
        """
        :type root: Node
        :rtype: List[int]
        """
        if not root: return []
        st=[root]
        res=[]
        while st:
            r=st.pop()
            res.append(r.val)
            st += r.children[::-1]
        return res
        

class Solution(object):
    def postorder(self, root):
        """
        :type root: Node
        :rtype: List[int]
        """
        if not root: return []
        res, st = [], [root]
        while st:
            r=st.pop()
            res.append(r.val)
            st+=r.children
        return res[::-1]

这种写法也是super simple啊