老师，我想求出二叉树中两个节点的距离，我的代码如下，函数已经写出来了，但是不知道如何返回level值，麻烦您帮我改一下，谢谢您了
class BiTNode:
def __init__(self):
self.data = None
self.lchild = None
self.rchild = None
def arraytotree(arr,start,end):
#中序遍历生成二叉树
root = None
if end >= start:
root = BiTNode()
mid = (start+end+1)//2
root.data = arr[mid]
root.lchild = arraytotree(arr,start,mid-1)
root.rchild = arraytotree(arr,mid+1,end)
else:
root = None
return root
def FindParentNode(root,node1,node2):
#找到公共父节点
if root == None or root == node1 or root == node2:
return root
lchild = FindParentNode(root.lchild,node1,node2)
rchild = FindParentNode(root.rchild,node1,node2)
if lchild == None:
return rchild
elif rchild == None:
return lchild
else: return root
def printTree(root,node,level=1):
#输出节点与根节点的距离level
if root == node:
return 0
if root.lchild != None:
if root.lchild == node:
print(level)
printTree(root.lchild,node,level=level+1)
# print(root.data,end=" ")
if root.rchild != None:
if root.rchild == node:
print(level)
printTree(root.rchild,node,level=level+1)
if __name__=="__main__":
arr = [1,2,3,4,5,6,7,8,9,10]
root = arraytotree(arr,0,len(arr)-1)
node1 = root.rchild.lchild.lchild
node2 = root.lchild.rchild.lchild
L = printTree(root,node1)
M = printTree(root,node2)
res = FindParentNode(root,node1,node2)
N = printTree(root,res)
print(L,M,N)
distance = printTree(root,node1)+printTree(root,node2)-2*printTree(root,res)
# if res != None:
# print(str(node1.data))
# print(str(node2.data))
# print(str(res.data))
# print(distance)
两种影响的时候给这个父元素加clear:both;有意义吗，这个不是加在受影响的元素里面吗
老师这是什么原因呀，那个平均薪资打印不出来
def哪错了 #哪错了aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
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