FBI树【python】【顺序存储二叉树解法】

解题思路:
顺序存储二叉树
注意事项:

参考代码:

n=int(input())
s=input().strip()
tree=['']*100000 # 存储每个节点的子串


def fun(tree,node,s):#构建树
    start=0
    end=len(s)-1
    #print(s)
    tree[node]=s
    if start==end:
        return
    mid=(start+end)//2
    left_node=2*node+1
    right_node=2*node+2
    fun(tree,left_node,s[start:mid+1])
    fun(tree,right_node,s[mid+1:end+1])


fun(tree,0,s,)


#print(tree)
def check(s):   #根据字串返回节点
    #print(s,end=' ')
    b=0
    I=0
    for i in range(len(s)):
        if s[i]=='0':
            b+=1
        else:
            I+=1
    if b==0:
        return 'I'
    if I==0:
        return 'B'
    return 'F'

def postOrderTree(tree,node):#后序遍历

    left_node=2*node+1
    right_node=2*node+2
    if len(tree[node])!=1:
        postOrderTree(tree,left_node)
    if len(tree[node])!=1:
        postOrderTree(tree,right_node)
    print(check(tree[node]),end='')
    #print(tree[node])
postOrderTree(tree,0)