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leetcode-problemset/leetcode-cn/originData/P5rCT8.json
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"question": {
"questionId": "1000313",
"questionFrontendId": "剑指 Offer II 053",
"categoryTitle": "LCOF2",
"boundTopicId": 910307,
"title": "二叉搜索树中的中序后继",
"titleSlug": "P5rCT8",
"content": "<p>English description is not available for the problem. Please switch to Chinese.</p>\n",
"translatedTitle": "二叉搜索树中的中序后继",
"translatedContent": "<p>给定一棵二叉搜索树和其中的一个节点 <code>p</code> ,找到该节点在树中的中序后继。如果节点没有中序后继,请返回 <code>null</code> 。</p>\n\n<p>节点&nbsp;<code>p</code>&nbsp;的后继是值比&nbsp;<code>p.val</code>&nbsp;大的节点中键值最小的节点,即按中序遍历的顺序节点 <code>p</code> 的下一个节点。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/01/23/285_example_1.PNG\" style=\"height: 117px; width: 122px;\" /></p>\n\n<pre>\n<strong>输入:</strong>root = [2,1,3], p = 1\n<strong>输出:</strong>2\n<strong>解释:</strong>这里 1 的中序后继是 2。请注意 p 和返回值都应是 TreeNode 类型。\n</pre>\n\n<p><strong>示例&nbsp;2</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/01/23/285_example_2.PNG\" style=\"height: 229px; width: 246px;\" /></p>\n\n<pre>\n<strong>输入:</strong>root = [5,3,6,2,4,null,null,1], p = 6\n<strong>输出:</strong>null\n<strong>解释:</strong>因为给出的节点没有中序后继,所以答案就返回 <code>null 了。</code>\n</pre>\n\n<p>&nbsp;</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>[1, 10<sup>4</sup>]</code> 内。</li>\n\t<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>\n\t<li>树中各节点的值均保证唯一。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><meta charset=\"UTF-8\" />注意:本题与主站 285&nbsp;题相同:&nbsp;<a href=\"https://leetcode-cn.com/problems/inorder-successor-in-bst/\">https://leetcode-cn.com/problems/inorder-successor-in-bst/</a></p>\n",
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"code": "/**\n * Definition for a binary tree node.\n * struct TreeNode {\n * int val;\n * TreeNode *left;\n * TreeNode *right;\n * TreeNode(int x) : val(x), left(NULL), right(NULL) {}\n * };\n */\nclass Solution {\npublic:\n TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {\n \n }\n};",
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"code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * int val;\n * TreeNode left;\n * TreeNode right;\n * TreeNode(int x) { val = x; }\n * }\n */\nclass Solution {\n public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {\n \n }\n}",
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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':\n ",
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"code": "/**\n * Definition for a binary tree node.\n * function TreeNode(val) {\n * this.val = val;\n * this.left = this.right = null;\n * }\n */\n/**\n * @param {TreeNode} root\n * @param {TreeNode} p\n * @return {TreeNode}\n */\nvar inorderSuccessor = function(root, p) {\n \n};",
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"code": "/**\n * Definition for a binary tree node.\n * public class TreeNode {\n * public var val: Int\n * public var left: TreeNode?\n * public var right: TreeNode?\n * public init(_ val: Int) {\n * self.val = val\n * self.left = nil\n * self.right = nil\n * }\n * }\n */\n\nclass Solution {\n func inorderSuccessor(_ root: TreeNode?, _ p: TreeNode?) -> TreeNode? {\n \n }\n}",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode(var `val`: Int = 0) {\n * var left: TreeNode? = null\n * var right: TreeNode? = null\n * }\n */\n\nclass Solution {\n fun inorderSuccessor(root: TreeNode?, p: TreeNode?): TreeNode? {\n \n }\n}",
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