分類：二叉樹-二叉樹的遍歷

題目描述：給定一個二叉樹，返回它的後序遍歷。

解題思路：

遞迴：

# 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 postorderTraversal(self, root: TreeNode) -> List[int]:
        遞迴
        
 
def postorder(root:TreeNode):
            if not root:
                return []
            postorder(root.left)
            postorder(root.right)
            res.append(root.val)
        res = []
        postorder(root)
        return res

迭代：

# 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 postorderTraversal(self, root: TreeNode) -> List[int]:
       # 迭代
        if not root:
            return []
        res,stack = [],[root]
        prev 
 
= root # 為了判斷父子節點關係
        while len(stack):
            root = stack.pop()# 取出一個節點，表示開始訪問以該節點為根的子樹
            if (not root.right and not root.left ) or ( root.right ==prev or root.left == prev): 
                # 如果該節點為葉子節點，或者已經訪問該節點的子節點
                res.append(root.val) # 直接訪問
                prev = root
            else:# 否則就順序把當前節點，右孩子，左孩子入棧
                stack.append(root)
                if root.right:
                    stack.append(root.right)
                if root.left:
                    stack.append(root.left)
        return res

# 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 postorderTraversal(self, root: TreeNode) -> List[int]:
        if not root:
            return []
        res,stack=[],[]
        prev = None
        while root or stack:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            # 比前序和中序的模板增加一個判斷過程：節點沒有右孩子或已經訪問了該節點的子節點
            if not root.right or root.right == prev:
                res.append(root.val)
                prev = root
                root = None 
            else:
                stack.append(root)
                root = root.right
        return res