层序遍历

层序遍历一个二叉树。就是从左到右一层一层的去遍历二叉树。

需要借用一个辅助数据结构即队列来实现，

队列先进先出，符合一层一层遍历的逻辑，

而用栈先进后出适合模拟深度优先遍历也就是递归的逻辑。

而这种层序遍历方式就是图论中的广度优先遍历

102. 二叉树的层序遍历

层序遍历难点在于，获取每一层的值，那么我们使用队列假设是q仅用来处理每层的元素即可，即for i in len（q），队列我们使用可左弹出的collection.deque

# 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]]:
        que=collections.deque([root])
        ret = []
        if root is None:
            return []
        while que:
            level = []
            for i in range(len(que)):
                cur = que.popleft()
                level.append(cur.val)
                if cur.left:
                    que.append(cur.left)
                if cur.right:
                    que.append(cur.right)
            ret.append(level)
        return ret

226. 翻转二叉树

这道题目背后有一个让程序员心酸的故事，听说 Homebrew的作者Max Howell，就是因为没在白板上写出翻转二叉树，最后被Google拒绝了。（真假不做判断，权当一个乐子哈）

递归法：前序遍历：

# 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 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

迭代法：前序遍历：

# 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 invertTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return None      
        stack = [root]        
        while stack:
            node = stack.pop()   
            node.left, node.right = node.right, node.left                   
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)  
        return root

递归法：中序遍历：

# 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 invertTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return None
        self.invertTree(root.left)
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        return root

迭代法：中序遍历：

# 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 invertTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return None      
        stack = [root]        
        while stack:
            node = stack.pop()                   
            if node.left:
                stack.append(node.left)
            node.left, node.right = node.right, node.left               
            if node.left:
                stack.append(node.left)       
        return root

递归法：后序遍历：

# 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 invertTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return None
        self.invertTree(root.left)
        self.invertTree(root.right)
        root.left, root.right = root.right, root.left
        return root

迭代法：后序遍历：

# 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 invertTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return None      
        stack = [root]        
        while stack:
            node = stack.pop()                   
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)  
            node.left, node.right = node.right, node.left               
     
        return root

迭代法：广度优先遍历（层序遍历）：

# 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 invertTree(self, root: TreeNode) -> TreeNode:
        if not root: 
            return None

        queue = collections.deque([root])    
        while queue:
            for i in range(len(queue)):
                node = queue.popleft()
                node.left, node.right = node.right, node.left
                if node.left: queue.append(node.left)
                if node.right: queue.append(node.right)
        return root