已知二叉树前序和中序遍历，求后序遍历

shen464165094

网上搜到的代码
pre = [3, 9, 20, 15, -6, 7]
tin = [9, 3, -6, 15, 20, 7]
global aftree
global ai
aftree = []
ai = 0
def reConstruction(pre,tin):
    print(pre,tin,sep=' ')
    if len(pre) == 0:
        return 'a'
    if len(pre) == 1:
        return pre[0]
    aftree.append(pre[0])
    print(aftree,'0',sep=' ')
    i = 0
    for i in range(len(tin)):
        if tin[i] == pre[0]:
            break
    left_pre = pre[1:1+i]
    right_pre = pre[1+i:]
    left_tin = tin[:i]
    right_tin = tin[i+1:]
    aftree.append(reConstruction(right_pre,right_tin))
    print(aftree,'r',sep=' ')
    aftree.append(reConstruction(left_pre,left_tin))
    print(aftree,'l',sep=' ')
   
    print('ai')

reConstruction(pre,tin)

tmp = []
for i in aftree:
    if i != None:
        tmp.append(i)

#res = ''.join(str(tmp))
print(tmp[::-1])



实在搞不懂，aftree中[3, 20, 7, 15, 'a', -6, None, None, 9]，这两个none怎么来的？？

shen464165094

想起来了  没有return  就是return none

陈尚涵

既然问题解决，那就设置个最佳答案吧~