{ "data": { "question": { "questionId": "94", "questionFrontendId": "94", "categoryTitle": "Algorithms", "boundTopicId": 1116, "title": "Binary Tree Inorder Traversal", "titleSlug": "binary-tree-inorder-traversal", "content": "
Given the root
of a binary tree, return the inorder traversal of its nodes' values.
\n
Example 1:
\n\n\nInput: root = [1,null,2,3]\nOutput: [1,3,2]\n\n\n
Example 2:
\n\n\nInput: root = []\nOutput: []\n\n\n
Example 3:
\n\n\nInput: root = [1]\nOutput: [1]\n\n\n
\n
Constraints:
\n\n[0, 100]
.-100 <= Node.val <= 100
\nFollow up: Recursive solution is trivial, could you do it iteratively?", "translatedTitle": "二叉树的中序遍历", "translatedContent": "
给定一个二叉树的根节点 root
,返回 它的 中序 遍历 。
\n\n
示例 1:
\n\n\n输入:root = [1,null,2,3]\n输出:[1,3,2]\n\n\n
示例 2:
\n\n\n输入:root = []\n输出:[]\n\n\n
示例 3:
\n\n\n输入:root = [1]\n输出:[1]\n\n\n
\n\n
提示:
\n\n[0, 100]
内-100 <= Node.val <= 100
\n\n
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
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