1、假设二叉树中的每个结点值为单个字符，采用顺序存储结构存储，设计一个算法，求二叉树t中的叶子结点个数。
2、假设二叉树以二叉链存储，设计一个算法，判断一个二叉树是否为完全二叉树。
用python做出来，分为两个py文件

我答第二题
class Node(object):
    def __init__(self, elem=-1, lchild=None, rchild=None):
        self.elem = elem
        self.lchild = lchild
        self.rchild = rchild

def trans(a,i,n): #n是以顺序存储元素个数
    if (i>n):
        return None                  
    b=Node(a[i-1])   
    if b.elem=='#':
        return None
    b.lchild=trans(a,2*i,n)
    b.rchild=trans(a,2*i+1,n)            
    return b

def isCBTree(root):   
    if not root:
        return False
    queue = [root]
    result = True                                
    hasNoChild=False
    while queue:
        cur=queue.pop(0)
        if hasNoChild:
            if cur.lchild or cur.rchild:
                return False
                break
        else:
            if cur.lchild and cur.rchild:
                queue.append(cur.lchild)
                queue.append(cur.rchild)
            elif cur.lchild and not cur.rchild:
                queue.append(cur.lchild)
                hasNoChild=True
            elif not cur.lchild and cur.rchild:
                result=False
                break
            else:
                hasNoChild=True
    return result



if __name__=='__main__':
    a=input('请以顺序存储的方式输入二叉树，用#补成满二叉树：')
    n=len(a)
    cc=isCBTree(trans(a,1,n))
    if cc==True:
        print('此二叉树是完全二叉树')
    else:
        print('此二叉树不是完全二叉树')

我答第二题
class Node(object):
    def __init__(self, elem=-1, lchild=None, rchild=None):
        self.elem = elem
        self.lchild = lchild
        self.rchild = rchild

def trans(a,i,n): #n是以顺序存储元素个数
    if (i>n):
        return None                  
    b=Node(a[i-1])   
    if b.elem=='#':
        return None
    b.lchild=trans(a,2*i,n)
    b.rchild=trans(a,2*i+1,n)            
    return b

def isCBTree(root):   
    if not root:
        return False
    queue = [root]
    result = True                                
    hasNoChild=False
    while queue:
        cur=queue.pop(0)
        if hasNoChild:
            if cur.lchild or cur.rchild:
                return False
                break
        else:
            if cur.lchild and cur.rchild:
                queue.append(cur.lchild)
                queue.append(cur.rchild)
            elif cur.lchild and not cur.rchild:
                queue.append(cur.lchild)
                hasNoChild=True
            elif not cur.lchild and cur.rchild:
                result=False
                break
            else:
                hasNoChild=True
    return result



if __name__=='__main__':
    a=input('请以顺序存储的方式输入二叉树，用#补成满二叉树：')
    n=len(a)
    cc=isCBTree(trans(a,1,n))
    if cc==True:
        print('此二叉树是完全二叉树')
    else:
        print('此二叉树不是完全二叉树')