145.binary-tree-postorder-traversal

Statement

简体中文English

Metadata

• Link: 二叉树的后序遍历
• Difficulty: Easy
• Tag: 栈 树 深度优先搜索 二叉树

给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。

 

示例 1：

输入：root = [1,null,2,3]
输出：[3,2,1]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1]
输出：[1]

 

提示：

• 树中节点的数目在范围 [0, 100] 内
• -100 <= Node.val <= 100

 

进阶：递归算法很简单，你可以通过迭代算法完成吗？

Metadata

• Link: Binary Tree Postorder Traversal
• Difficulty: Easy
• Tag: Stack Tree Depth-First Search Binary Tree

Given the root of a binary tree, return the postorder traversal of its nodes' values.

 

Example 1:

Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

 

Constraints:

• The number of the nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

 

Follow up: Recursive solution is trivial, could you do it iteratively?

Solution

Python 3

# 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
from typing import List, Optional


class Solution:
    def __init__(self):
        self.res = []

    def dfs(self, rt: TreeNode) -> None:
        if not rt:
            return

        self.dfs(rt.left)
        self.dfs(rt.right)
        self.res.append(rt.val)

    def postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        self.dfs(root)
        return self.res

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最后更新: October 11, 2023