原文連結:https://www.keketec.club/posts/11228680/
每天分享一個LeetCode題目
每天 5 分鐘,一起進步
LeetCode N 叉樹的後序周遊,位址: https://leetcode-cn.com/problems/n-ary-tree-postorder-traversal/
樹結點類
class TreeNode(object):
def __init__(self, val, children=[]):
self.val = val
self.children = children
N 叉樹的後序周遊
利用遞歸,依然遵循「左右根」的周遊原則
def postorder(self, root):
if not root:
return
for node in root.children:
self.postorder(node)
print(root.val, end=" ")
是不是看起來特别簡單,但是這樣不符合 LeetCode 題目中的要求
需要将結果放到一個數組中,是以需要提前初始化一個 list 進行存放
def postorder_lc(self, root):
res = []
def post_order(root):
if not root:
return
for node in root.children:
post_order(node)
res.append(root.val)
post_order(root)
return res
完整代碼
# -*- coding:utf-8 -*-
# !/usr/bin/env python
# 樹結點類
class Node(object):
def __init__(self, val=None, children=[]):
self.val = val
self.children = children
class Solution(object):
def postorder(self, root):
if not root:
return
for node in root.children:
self.postorder(node)
print(root.val, end=" ")
def postorder_lc(self, root):
res = []
def post_order(root):
if not root:
return
for node in root.children:
post_order(node)
res.append(root.val)
post_order(root)
return res
if __name__ == "__main__":
# 建立節點
root = Node('A')
node_B = Node('B')
node_C = Node('C')
node_D = Node('D')
node_E = Node('E')
node_F = Node('F')
node_G = Node('G')
node_H = Node('H')
node_I = Node('I')
# 建構三叉樹
# A
# / | \
# B C D
# /|\ / \
# E F G H I
root.children = [node_B, node_C, node_D]
node_B.children = [node_E, node_F, node_G]
node_D.children = [node_H, node_I]
s = Solution()
s.postorder(root)
print("\n")
print(s.postorder_lc(root))