145. Binary Tree Postorder Traversal

Problem

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

https://leetcode.com/problems/binary-tree-postorder-traversal/

Example 1:

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

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]
Output: [4,6,7,5,2,9,8,3,1]
Explanation:

Example 3:

Input: root = []
Output: []

Example 4:

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?

Test Cases

# 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: Optional[TreeNode]) -> List[int]:

solution_test.py

import pytest

import os
import sys
sys.path.append(os.path.dirname(os.path.dirname(__file__)))
from _utils.binary_tree import build_tree
from solution import Solution

null = None


@pytest.mark.parametrize('root, expected', [
    ([1,null,2,3], [3,2,1]),
    ([1,2,3,4,5,null,8,null,null,6,7,9], [4,6,7,5,2,9,8,3,1]),
    ([], []),
    ([1], [1]),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, root, expected):
    assert sol.postorderTraversal(build_tree(root)) == expected

Thoughts

在 124. Binary Tree Maximum Path Sum 或 337. House Robber III 中都用到了基于栈的非递归后序遍历二叉树。

Code

solution.py

from typing import Optional


# 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: Optional[TreeNode]) -> list[int]:
        res = []
        stack = []
        prev = None
        while root or stack:
            if root:
                stack.append(root)
                root = root.left
            elif stack[-1].right != prev:
                root = stack[-1].right
                prev = None
            else:
                prev = stack.pop()
                res.append(prev.val)

        return res