每日leetcode-数组-145. 二叉树的后序遍历

分类：二叉树-二叉树的遍历

题目描述：给定一个二叉树，返回它的 后序 遍历。

解题思路：

递归：

# 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]:
        递归
        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 # 为了判断父子节点关系
        while len(stack):
            root = stack.pop()# 取出一个节点，表示开始访问以该节点为根的子树
            if (not root.right and not root.left ) or ( root.right ==prev or root.left == prev): 
                # 如果该节点为叶子节点，或者已经访问该节点的子节点
                res.append(root.val) # 直接访问
                prev = root
            else:# 否则就顺序把当前节点，右孩子，左孩子入栈
                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个判断过程：节点没有右孩子或已经访问了该节点的子节点
            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