Day15 | 102. 二叉树的层序遍历 、226.翻转二叉树 101. 对称二叉树

102. 二叉树的层序遍历

看完本篇可以一口气刷十道题，试一试， 层序遍历并不难，大家可以很快刷了十道题。

题目链接/文章讲解/视频讲解：https://programmercarl.com/0102.二叉树的层序遍历.html

# 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 root is None:
            return []
        result = []
        queue = deque()
        queue.append(root)
        while queue:
            level = []
            for i in range(len(queue)):
                node = queue.popleft()
                level.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            result.append(level)
        return result

226.翻转二叉树 （优先掌握递归）

这道题目 一些做过的同学 理解的也不够深入，建议大家先看我的视频讲解，无论做过没做过，都会有很大收获。

题目链接/文章讲解/视频讲解：https://programmercarl.com/0226.翻转二叉树.html

# 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
def invert(root):
    if root is None:
        return 0
    root.left,root.right = root.right,root.left
    invert(root.left)
    invert(root.right)
class Solution:
    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        invert(root)
        return root

101. 对称二叉树 （优先掌握递归）

先看视频讲解，会更容易一些。

题目链接/文章讲解/视频讲解：https://programmercarl.com/0101.对称二叉树.html

# 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
def is_Symmetric(left,right):
    if (left is None and right is not None) or (left is not None and right is None):
        return False
    if left is None and right is None:
        return True    
    if left.val == right.val:
        return is_Symmetric(left.left,right.right) and is_Symmetric(left.right,right.left) 
    else:
        return False
    
class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        if root is None:
            return True
        return is_Symmetric(root.left,root.right)