代码随想录算法训练营第十五天| 层序遍历(写了三题)、226.翻转二叉树、101.对称二叉树
今天是代码随想录算法训练营第十五天
写了3道层次遍历的题目（应当还有7道可以照猫画虎）、226.翻转二叉树、101.对称二叉树
3道层次遍历的题目如下：
102. 二叉树的层序遍历
# 利用长度法
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
if not root:
return []
queue = collections.deque([root])
result = []
while queue:
level = []
for _ in range(len(queue)):
cur = queue.popleft()
level.append(cur.val)
if cur.left:
queue.append(cur.left)
if cur.right:
queue.append(cur.right)
result.append(level)
return result
- 二叉树的层序遍历 II
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
queue = collections.deque([root])
result = []
while queue:
level = []
for _ in range(len(queue)):
cur = queue.popleft()
level.append(cur.val)
if cur.left:
queue.append(cur.left)
if cur.right:
queue.append(cur.right)
result.append(level)
return result[::-1] #当start和stop都被省略时，切片将包含整个序列。而当step为负数时，切片将按相反的顺序提取序列中的元素，实现反转效果。
- 二叉树的右视图
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
if not root:
return []
queue = collections.deque([root])
right_view = []
while queue:
level_size = len(queue)
for i in range(level_size):
node = queue.popleft()
if i == level_size - 1:
right_view.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return right_view
226.翻转二叉树
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
# 这个Solution用了前序遍历（递归法），后序也是可以的，不过中序遍历会有点绕
class Solution:
def invertTree(self, root: TreeNode) -> TreeNode:
if not root:
return None
root.left, root.right = root.right, root.left
self.invertTree(root.left)
self.invertTree(root.right)
return root
101.对称二叉树
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
# 特别注意这里是需要使用后序遍历的
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
return self.compare(root.left, root.right)
def compare(self, left, right):
#首先排除空节点的情况
if left == None and right != None: return False
elif left != None and right == None: return False
elif left == None and right == None: return True
#排除了空节点，再排除数值不相同的情况
elif left.val != right.val: return False
#此时就是：左右节点都不为空，且数值相同的情况
#此时才做递归，做下一层的判断
# 特别注意这里是需要使用后序遍历的
outside = self.compare(left.left, right.right) #左子树：左、 右子树：右
inside = self.compare(left.right, right.left) #左子树：右、 右子树：左
isSame = outside and inside #左子树：中、 右子树：中 （逻辑处理）
return isSame