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        "question": {
            "questionId": "1114",
            "questionFrontendId": "1038",
            "categoryTitle": "Algorithms",
            "boundTopicId": 3836,
            "title": "Binary Search Tree to Greater Sum Tree",
            "titleSlug": "binary-search-tree-to-greater-sum-tree",
            "content": "<p>Given the <code>root</code> of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.</p>\n\n<p>As a reminder, a <em>binary search tree</em> is a tree that satisfies these constraints:</p>\n\n<ul>\n\t<li>The left subtree of a node contains only nodes with keys <strong>less than</strong> the node&#39;s key.</li>\n\t<li>The right subtree of a node contains only nodes with keys <strong>greater than</strong> the node&#39;s key.</li>\n\t<li>Both the left and right subtrees must also be binary search trees.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/05/02/tree.png\" style=\"width: 400px; height: 273px;\" />\n<pre>\n<strong>Input:</strong> root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]\n<strong>Output:</strong> [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [0,null,1]\n<strong>Output:</strong> [1,null,1]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is in the range <code>[1, 100]</code>.</li>\n\t<li><code>0 &lt;= Node.val &lt;= 100</code></li>\n\t<li>All the values in the tree are <strong>unique</strong>.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Note:</strong> This question is the same as 538: <a href=\"https://leetcode.com/problems/convert-bst-to-greater-tree/\" target=\"_blank\">https://leetcode.com/problems/convert-bst-to-greater-tree/</a></p>\n",
            "translatedTitle": "从二叉搜索树到更大和树",
            "translatedContent": "<p><span style=\"font-size:10.5pt\"><span style=\"font-family:Calibri\"><span style=\"font-size:10.5000pt\"><span style=\"font-family:宋体\"><font face=\"宋体\">给定一个二叉搜索树</font></span></span></span></span>&nbsp;<code>root</code>&nbsp;(BST)<span style=\"font-size:10.5pt\"><span style=\"font-family:Calibri\"><span style=\"font-size:10.5000pt\"><span style=\"font-family:宋体\"><font face=\"宋体\">，请将它的每个</font></span></span></span></span>节点<span style=\"font-size:10.5pt\"><span style=\"font-family:Calibri\"><span style=\"font-size:10.5000pt\"><span style=\"font-family:宋体\"><font face=\"宋体\">的值替换成树中大于或者等于该</font></span></span></span></span>节点<span style=\"font-size:10.5pt\"><span style=\"font-family:Calibri\"><span style=\"font-size:10.5000pt\"><span style=\"font-family:宋体\"><font face=\"宋体\">值的所有</font></span></span></span></span>节点<span style=\"font-size:10.5pt\"><span style=\"font-family:Calibri\"><span style=\"font-size:10.5000pt\"><span style=\"font-family:宋体\"><font face=\"宋体\">值之和。</font></span></span></span></span></p>\n\n<p>提醒一下， <em>二叉搜索树</em> 满足下列约束条件：</p>\n\n<ul>\n\t<li>节点的左子树仅包含键<strong> 小于 </strong>节点键的节点。</li>\n\t<li>节点的右子树仅包含键<strong> 大于</strong> 节点键的节点。</li>\n\t<li>左右子树也必须是二叉搜索树。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/05/03/tree.png\" style=\"height:273px; width:400px\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>[4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]\n<strong>输出：</strong>[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [0,null,1]\n<strong>输出：</strong>[1,null,1]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中的节点数在&nbsp;<code>[1, 100]</code>&nbsp;范围内。</li>\n\t<li><code>0 &lt;= Node.val &lt;= 100</code></li>\n\t<li>树中的所有值均 <strong>不重复</strong>&nbsp;。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>注意：</strong>该题目与 538:&nbsp;<a href=\"https://leetcode-cn.com/problems/convert-bst-to-greater-tree/\">https://leetcode-cn.com/problems/convert-bst-to-greater-tree/&nbsp; </a>相同</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() : val(0), left(nullptr), right(nullptr) {}\n *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}\n *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}\n * };\n */\nclass Solution {\npublic:\n    TreeNode* bstToGst(TreeNode* root) {\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() {}\n *     TreeNode(int val) { this.val = val; }\n *     TreeNode(int val, TreeNode left, TreeNode right) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\nclass Solution {\n    public TreeNode bstToGst(TreeNode root) {\n\n    }\n}",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode(object):\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution(object):\n    def bstToGst(self, root):\n        \"\"\"\n        :type root: TreeNode\n        :rtype: TreeNode\n        \"\"\"\n        ",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\nclass Solution:\n    def bstToGst(self, root: TreeNode) -> TreeNode:\n        ",
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                    "lang": "C",
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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* bstToGst(struct TreeNode* root){\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 val=0, TreeNode left=null, TreeNode right=null) {\n *         this.val = val;\n *         this.left = left;\n *         this.right = right;\n *     }\n * }\n */\npublic class Solution {\n    public TreeNode BstToGst(TreeNode root) {\n\n    }\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * function TreeNode(val, left, right) {\n *     this.val = (val===undefined ? 0 : val)\n *     this.left = (left===undefined ? null : left)\n *     this.right = (right===undefined ? null : right)\n * }\n */\n/**\n * @param {TreeNode} root\n * @return {TreeNode}\n */\nvar bstToGst = function(root) {\n\n};",
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                    "code": "# Definition for a binary tree node.\n# class TreeNode\n#     attr_accessor :val, :left, :right\n#     def initialize(val = 0, left = nil, right = nil)\n#         @val = val\n#         @left = left\n#         @right = right\n#     end\n# end\n# @param {TreeNode} root\n# @return {TreeNode}\ndef bst_to_gst(root)\n\nend",
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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() { self.val = 0; self.left = nil; self.right = nil; }\n *     public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }\n *     public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {\n *         self.val = val\n *         self.left = left\n *         self.right = right\n *     }\n * }\n */\nclass Solution {\n    func bstToGst(_ root: TreeNode?) -> TreeNode? {\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 bstToGst(root *TreeNode) *TreeNode {\n\n}",
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                    "lang": "Rust",
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                    "code": "// Definition for a binary tree node.\n// #[derive(Debug, PartialEq, Eq)]\n// pub struct TreeNode {\n//   pub val: i32,\n//   pub left: Option<Rc<RefCell<TreeNode>>>,\n//   pub right: Option<Rc<RefCell<TreeNode>>>,\n// }\n//\n// impl TreeNode {\n//   #[inline]\n//   pub fn new(val: i32) -> Self {\n//     TreeNode {\n//       val,\n//       left: None,\n//       right: None\n//     }\n//   }\n// }\nuse std::rc::Rc;\nuse std::cell::RefCell;\nimpl Solution {\n    pub fn bst_to_gst(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {\n\n    }\n}",
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                    "code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n *     val: number\n *     left: TreeNode | null\n *     right: TreeNode | null\n *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {\n *         this.val = (val===undefined ? 0 : val)\n *         this.left = (left===undefined ? null : left)\n *         this.right = (right===undefined ? null : right)\n *     }\n * }\n */\n\nfunction bstToGst(root: TreeNode | null): TreeNode | null {\n\n};",
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