前序

# recursively
def preorderTraversal1(self, root):
    res = []
    self.dfs(root, res)
    return res
    
def dfs(self, root, res):
    if root:
        res.append(root.val)
        self.dfs(root.left, res)
        self.dfs(root.right, res)

# iteratively
class Solution:
    def preorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        stack,ans=[root],[]
        while stack:
            node=stack.pop()
            ans.append(node.val)
            if node.right:
                stack.append(node.right)
            if node.left:
                stack.append(node.left)
        return ans

中序

# recursively
def inorderTraversal1(self, root):
    res = []
    self.helper(root, res)
    return res
    
def helper(self, root, res):
    if root:
        self.helper(root.left, res)
        res.append(root.val)
        self.helper(root.right, res)
 

# iteratively 
class Solution:
    def inorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        cur,stack,ans=root,[],[]
        while stack or cur:
            while cur:
                stack.append(cur)
                cur=cur.left
            cur=stack.pop()
            ans.append(cur.val)
            cur=cur.right
        return ans

后序

# recursively 
def postorderTraversal1(self, root):
    res = []
    self.dfs(root, res)
    return res
    
def dfs(self, root, res):
    if root:
        self.dfs(root.left, res)
        self.dfs(root.right, res)
        res.append(root.val)
        
## 标准的后序迭代
class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        res = []   # 用来存储后序遍历节点的值
        stack = []  
        node = root
        while stack or node:
            while node:
                stack.append(node)  # 第一次入栈的是根节点
                #判断当前节点的左子树是否存在，若存在则持续左下行，若不存在就转向右子树
                node = node.left if node.left is not None else node.right
            #循环结束说明走到了叶子节点，没有左右子树了，该叶子节点即为当前栈顶元素，应该访问了
            node = stack.pop() # 取出栈顶元素进行访问
            res.append(node.val) # 将栈顶元素也即当前节点的值添加进res
            # （下面的stack[-1]是执行完上面那句取出栈顶元素后的栈顶元素）
            if stack and stack[-1].left == node: #若栈不为空且当前节点是栈顶元素的左节点
                node = stack[-1].right   ## 则转向遍历右节点
            else:
                node = None  # 没有左子树或右子树，强迫退栈
        return res        
       
        
# iteratively2 保存节点遍历状态的迭代，和使用双栈，一个栈放node,一个栈放flag一样
class Solution:
    def postorderTraversal(self, root):
        traversal, stack = [], [(root, False)]
        while stack:
            node, visited = stack.pop()
            if node:
                if visited:
                    # add to result if visited
                    traversal.append(node.val)
                else:
                    # post-order
                    stack.append((node, True))
                    stack.append((node.right, False))
                    stack.append((node.left, False))
        return traversal
        
# iteratively3(前序遍历取反，最好不要用) 
def postorderTraversal(self, root):
    res, stack = [], [root]
    while stack:
        node = stack.pop()
        if node:
            res.append(node.val)
            stack.append(node.left)
            stack.append(node.right)
    return res[::-1]

层序

从第一层打印到最后一层

class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        ans=[]
        cur_stack=[root]
        while cur_stack:
            next_stack,tmp=[],[]
            for node in cur_stack:
                tmp.append(node.val)
                if node.left:
                    next_stack.append(node.left)
                if node.right:
                    next_stack.append(node.right)
            cur_stack=next_stack
            ## 如果是从最后一层打印到第一层，改成ans.insert(0,tmp)就好了
            ans.append(tmp)
        return ans