mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-09-02 22:13:28 +08:00
update
This commit is contained in:
@@ -12,7 +12,7 @@
|
||||
"translatedContent": "<p>给你一棵二叉树的根节点 <code>root</code> ,返回其节点值的 <strong>后序遍历 </strong>。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/08/28/pre1.jpg\" style=\"width: 127px; height: 200px;\" />\n<pre>\n<strong>输入:</strong>root = [1,null,2,3]\n<strong>输出:</strong>[3,2,1]\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>root = []\n<strong>输出:</strong>[]\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong>root = [1]\n<strong>输出:</strong>[1]\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>树中节点的数目在范围 <code>[0, 100]</code> 内</li>\n\t<li><code>-100 <= Node.val <= 100</code></li>\n</ul>\n\n<p> </p>\n\n<p><strong>进阶:</strong>递归算法很简单,你可以通过迭代算法完成吗?</p>\n",
|
||||
"isPaidOnly": false,
|
||||
"difficulty": "Easy",
|
||||
"likes": 794,
|
||||
"likes": 820,
|
||||
"dislikes": 0,
|
||||
"isLiked": null,
|
||||
"similarQuestions": "[{\"title\": \"Binary Tree Inorder Traversal\", \"titleSlug\": \"binary-tree-inorder-traversal\", \"difficulty\": \"Easy\", \"translatedTitle\": \"\\u4e8c\\u53c9\\u6811\\u7684\\u4e2d\\u5e8f\\u904d\\u5386\"}, {\"title\": \"N-ary Tree Postorder Traversal\", \"titleSlug\": \"n-ary-tree-postorder-traversal\", \"difficulty\": \"Easy\", \"translatedTitle\": \"N \\u53c9\\u6811\\u7684\\u540e\\u5e8f\\u904d\\u5386\"}]",
|
||||
@@ -155,7 +155,7 @@
|
||||
"__typename": "CodeSnippetNode"
|
||||
}
|
||||
],
|
||||
"stats": "{\"totalAccepted\": \"402.5K\", \"totalSubmission\": \"532.5K\", \"totalAcceptedRaw\": 402472, \"totalSubmissionRaw\": 532486, \"acRate\": \"75.6%\"}",
|
||||
"stats": "{\"totalAccepted\": \"423.6K\", \"totalSubmission\": \"559.6K\", \"totalAcceptedRaw\": 423613, \"totalSubmissionRaw\": 559598, \"acRate\": \"75.7%\"}",
|
||||
"hints": [],
|
||||
"solution": null,
|
||||
"status": null,
|
||||
|
Reference in New Issue
Block a user