leetcode-problemset/leetcode-cn/originData/successor-lcci.json

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        "question": {
            "questionId": "100178",
            "questionFrontendId": "面试题 04.06",
            "categoryTitle": "LCCI",
            "boundTopicId": 46213,
            "title": "Successor LCCI",
            "titleSlug": "successor-lcci",
            "content": "<p>Write an algorithm to find the &quot;next&quot; node (i.e., in-order successor) of a given node in a binary search tree.</p>\r\n\r\n<p>Return <code>null</code> if there&#39;s no &quot;next&quot; node for the given node.</p>\r\n\r\n<p><strong>Example 1:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> root = <code>[2,1,3], p = 1\r\n\r\n  2\r\n / \\\r\n1   3\r\n</code>\r\n<strong>Output:</strong> 2</pre>\r\n\r\n<p><strong>Example 2:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> root = <code>[5,3,6,2,4,null,null,1], p = 6\r\n\r\n      5\r\n     / \\\r\n    3   6\r\n   / \\\r\n  2   4\r\n /   \r\n1\r\n</code>\r\n<strong>Output:</strong> null</pre>\r\n",
            "translatedTitle": "后继者",
            "translatedContent": "<p>设计一个算法，找出二叉搜索树中指定节点的&ldquo;下一个&rdquo;节点（也即中序后继）。</p>\n\n<p>如果指定节点没有对应的&ldquo;下一个&rdquo;节点，则返回<code>null</code>。</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre><strong>输入:</strong> root = <code>[2,1,3], p = 1\n\n  2\n / \\\n1   3\n</code>\n<strong>输出:</strong> 2</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre><strong>输入:</strong> root = <code>[5,3,6,2,4,null,null,1], p = 6\n\n      5\n     / \\\n    3   6\n   / \\\n  2   4\n /   \n1\n</code>\n<strong>输出:</strong> null</pre>\n",
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            "topicTags": [
                {
                    "name": "Tree",
                    "slug": "tree",
                    "translatedName": "树",
                    "__typename": "TopicTagNode"
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                {
                    "name": "Depth-First Search",
                    "slug": "depth-first-search",
                    "translatedName": "深度优先搜索",
                    "__typename": "TopicTagNode"
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                {
                    "name": "Binary Search Tree",
                    "slug": "binary-search-tree",
                    "translatedName": "二叉搜索树",
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                    "name": "Binary Tree",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, x):\n#         self.val = x\n#         self.left = None\n#         self.right = None\n\nclass Solution:\n    def inorderSuccessor(self, root: TreeNode, p: TreeNode) -> TreeNode:",
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                    "code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n *     int val;\n *     struct TreeNode *left;\n *     struct TreeNode *right;\n * };\n */\n\n\nstruct TreeNode* inorderSuccessor(struct TreeNode* root, struct TreeNode* p){\n\n}\n",
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                    "code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n *     public int val;\n *     public TreeNode left;\n *     public TreeNode right;\n *     public TreeNode(int x) { val = x; }\n * }\n */\npublic class Solution {\n    public TreeNode InorderSuccessor(TreeNode root, TreeNode p) {\n\n    }\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n *     Val int\n *     Left *TreeNode\n *     Right *TreeNode\n * }\n */\nfunc inorderSuccessor(root *TreeNode, p *TreeNode) *TreeNode {\n\n}",
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            "hints": [
                "想想中序遍历是如何工作的，并尝试对其进行“逆向工程”。",
                "这只是逻辑方法中的一步：一个特定节点的后继节点是右子树的最左节点。如果没有右子树呢？"
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