mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
187 lines
27 KiB
JSON
187 lines
27 KiB
JSON
{
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"data": {
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"question": {
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"questionId": "1000315",
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"questionFrontendId": "LCR 054",
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"categoryTitle": "Algorithms",
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"boundTopicId": 910309,
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"title": "把二叉搜索树转换为累加树",
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"titleSlug": "w6cpku",
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"content": "<p>English description is not available for the problem. Please switch to Chinese.</p>\n",
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"translatedTitle": "把二叉搜索树转换为累加树",
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"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，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 style=\"margin:0pt 0pt 0.0001pt; text-align:justify\"> </p>\n\n<p>提醒一下，二叉搜索树满足下列约束条件：</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> </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: 364px; width: 534px;\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>root<strong> </strong>=<strong> </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><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [1,0,2]\n<strong>输出：</strong>[3,3,2]\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [3,2,4,1]\n<strong>输出：</strong>[7,9,4,10]\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中的节点数介于 <code>0</code> 和 <code>10<sup>4</sup></code><sup> </sup>之间。</li>\n\t<li>每个节点的值介于 <code>-10<sup>4</sup></code> 和 <code>10<sup>4</sup></code> 之间。</li>\n\t<li>树中的所有值 <strong>互不相同</strong> 。</li>\n\t<li>给定的树为二叉搜索树。</li>\n</ul>\n\n<p> </p>\n\n<p><meta charset=\"UTF-8\" />注意：</p>\n\n<ul>\n\t<li>本题与主站 538 题相同： <a href=\"https://leetcode-cn.com/problems/convert-bst-to-greater-tree/\">https://leetcode-cn.com/problems/convert-bst-to-greater-tree/</a></li>\n\t<li>本题与主站 1038 题相同：<a href=\"https://leetcode-cn.com/problems/binary-search-tree-to-greater-sum-tree/\">https://leetcode-cn.com/problems/binary-search-tree-to-greater-sum-tree/</a></li>\n</ul>\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 ConvertBST(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 convertBST = function(root) {\n\n};",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * public $val = null;\n * public $left = null;\n * public $right = null;\n * function __construct($val = 0, $left = null, $right = null) {\n * $this->val = $val;\n * $this->left = $left;\n * $this->right = $right;\n * }\n * }\n */\nclass Solution {\n\n /**\n * @param TreeNode $root\n * @return TreeNode\n */\n function convertBST($root) {\n\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() { 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 convertBST(_ root: TreeNode?) -> TreeNode? {\n\n }\n}",
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"code": "/**\n * Example:\n * var ti = TreeNode(5)\n * var v = ti.`val`\n * Definition for a binary tree node.\n * class TreeNode(var `val`: Int) {\n * var left: TreeNode? = null\n * var right: TreeNode? = null\n * }\n */\nclass Solution {\n fun convertBST(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 convertBST(root *TreeNode) *TreeNode {\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 convert_bst(root)\n\nend",
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"code": "/**\n * Definition for a binary tree node.\n * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {\n * var value: Int = _value\n * var left: TreeNode = _left\n * var right: TreeNode = _right\n * }\n */\nobject Solution {\n def convertBST(root: TreeNode): TreeNode = {\n\n }\n}",
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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 convert_bst(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {\n\n }\n}",
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