答案:二叉树比较【难度：2级】--景越训练营Python习题答案（含多种解法）

二叉树比较【难度：2级】:

答案1：

def compare(a, b):
    return a.val == b.val and compare(a.left, b.left) and compare(a.right, b.right) if a and b else a == b​

答案2：

# return True if the two binary trees rooted and a and b are equal in value and structure
# return False otherwise
def compare(a, b):
  if a is None or b is None:
    return a is b
  return a.val == b.val and compare(a.left, b.left) and compare(a.right, b.right)
​

答案3：

compare = lambda a, b: (
    a and b
    and
        a.val == b.val
            and compare(a.left , b.left )
            and compare(a.right, b.right)
    or
        a == b
)​

答案4：

# return True if the two binary trees rooted and a and b are equal in value and structure
# return False otherwise
def compare(a, b):
    l1, l2 = [],[]
    in_order(a,l1)
    in_order(b,l2)
    return (l1 == l2)

def in_order(node, list):
    if node == None:
        return
    if node.left != None:
        in_order(node.left, list)
    list.append(node.val)
    if node.right != None:
        in_order(node.right,list)
    
​

答案5：

def compare(a, b):
    return a and b and a.val == b.val and compare(a.left, b.left) and compare(a.right, b.right) \
        or a is b is None​

答案6：

# return True if the two binary trees rooted and a and b are equal in value and structure
# return False otherwise
def compare(a, b):
    if a is None and b is None:
        return True
    if a is not None and b is not None:
        if a.val != b.val: 
            return False
        else:
            return compare(a.left, b.left) and compare(a.right, b.right)
    return False
​

答案7：

# return True if the two binary trees rooted and a and b are equal in value and structure
# return False otherwise
def compare(a, b):
    if (a == None and b == None):
        return True
    if ((a == None) ^ (b == None)) or (a.val != b.val):
        return False
    return (compare(a.left, b.left) and compare(a.right, b.right))
​

答案8：

def compare(a, b):
    return True if a == b == None else False if (a == None and b != None) or (a != None and b == None) else compare(a.left, b.left) and compare(a.right, b.right) if a.val == b.val else False​

答案9：

from operator import eq

def tree_to_inorder_list(node):
    return tree_to_inorder_list(node.left) + [node.val] + tree_to_inorder_list(node.right) if node else []

def compare(a, b):
  return eq(*map(tree_to_inorder_list, (a, b)))

​

答案10：

# return True if the two binary trees rooted and a and b are equal in value and structure
# return False otherwise
def compare(a, b):
    if not a and not b:
        return True
    if a and not b:
        return False
    if not a and b:
        return False
    return a.val == b.val and compare(a.left, b.left) and compare(a.right, b.right)
    ​