力扣Hot100--94.二叉树的中序遍历、144.二叉树的前序遍历、145.二叉树的后序遍历

要求：给定一个二叉树的根节点 root ，返回它的中序遍历。
递归解法：

python 复制代码

# 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 inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        def traversal(node):
            if node is None:
                return 
            traversal(node.left)# 左
            result.append(node.val) # 中
            traversal(node.right) # 右
        traversal(root)
        return result

迭代解法：

python 复制代码

class Solution:
    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        stack = []
        current = root
        while current or stack:
            if current:
                stack.append(current)
                current = current.left
            else:
                current = stack.pop()
                result.append(current.val)
                current = current.right
        return result

要求：给你二叉树的根节点 root ，返回它节点值的前序遍历。
递归解法：

python 复制代码

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        def traversal(node):
            if node is None:
                return
            result.append(node.val)
            traversal(node.left)
            traversal(node.right)
        traversal(root)
        return result

迭代解法：

python 复制代码

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if root is None:  # 一定要检查根节点是否为空
            return []
            
        result = []
        stack = [root]
        while stack:
            node = stack.pop()
            result.append(node.val)  

            if node.right:  # 先右后左
                stack.append(node.right)
            if node.left:   # 左子节点后入栈，会先出栈
                stack.append(node.left)

要求：给你一棵二叉树的根节点 root ，返回其节点值的后序遍历。
递归解法：

python 复制代码

class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        def traversal(node):
            if node is None:
                return 
            traversal(node.left)
            traversal(node.right)
            result.append(node.val)
        traversal(root)
        return result

迭代解法：

python 复制代码

class Solution:
    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if root is None:
            return []
        result = []
        stack = [root]
        while stack:
            node = stack.pop()
            result.append(node.val)
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
        return result[::-1]