Description
Given 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).
Follow up:
Recursive solution is trivial, could you do it iteratively?
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 height of the n-ary tree is less than or equal to 1000
- The total number of nodes is between [0, 10^4]
分析
题目的意思是：求n叉树的后续遍历，后续遍历是左右中，所以在递归的时候等孩子节点遍历完以后，最后放入结果集合里面。迭代的方法需要用到栈，孩子节点从左到右进去，然后栈里从右到左出来，这样中右左进行遍历，最后逆序一下结果，就可以变成左右中了，哈哈哈。
代码
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children
"""
class Solution:
def solve(self,root,res):
if(root is None):
return
for child in root.children:
self.solve(child,res)
res.append(root.val)
def postorder(self, root: 'Node') -> List[int]:
res=[]
self.solve(root,res)
return res
参考文献
[LeetCode] Beats 98.7%, 8 line pythonic iterative solution
