二叉树拓展--三叉树的遍历、路径算法

最新推荐文章于 2023-03-26 22:29:01 发布

Allegro很愉悦

最新推荐文章于 2023-03-26 22:29:01 发布

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文章标签：数据结构二叉树多叉树

本文链接：https://blog.youkuaiyun.com/weixin_42095500/article/details/83001049

nodes = ['a',['b',['e',['k','#','#','#'],['l','#','#','#'],'#'],['f',['m','#','#','#'],'#','#'],'#'],\
['c',['g',['n',['t','#','#','#'],['u','#','#','#'],'#'],'#','#'],['h',['o',['v',['y','#','#','#'],\
['z','#','#','#'],'#'],'#','#'],['p',['w','#','#','#'],['x','#','#','#'],'#'],'#'],\
['i',['q','#','#','#'],'#','#']],['d',['j',['r','#','#','#'],['s','#','#','#'],'#'],'#','#']]
# Definition for a binary tree node.
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.mid = None
self.right = None
#用嵌套列表构造三叉树，返回根节点
def buildTree(List):
if List == '#':
return
#print(List[0],List)
val,left,mid,right = List[0],List[1],List[2],List[3]
root = TreeNode(val)