题解 | #输出二叉树的右视图#

超简洁的代码，无需重构二叉树，直接输出矩阵结果，我太优秀了，爱自己

class Solution:
res={}
def solve(self , xianxu , zhongxu ):
res1=[]
self.reconstruct(xianxu,zhongxu,0)
for key in range(len(self.res)):
res1.append(self.res[key][-1])
return res1
def reconstruct(self, xianxu, zhongxu,level):
head=xianxu[0]
split=zhongxu.index(head)
if split!=0:
self.reconstruct(xianxu[1:split+1],zhongxu[:split],level+1)
if split+1<len(zhongxu):
self.reconstruct(xianxu[split+1:],zhongxu[split+1:],level+1)
if level in self.res:
self.res[level].append(head)
else:
self.res[level]=[head]
else:
if level in self.res:
self.res[level].append(xianxu[0])
else:
self.res[level]=[xianxu[0]]
return