mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
192 lines
25 KiB
JSON
192 lines
25 KiB
JSON
{
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"questionId": "1000307",
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"questionFrontendId": "剑指 Offer II 050",
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"categoryTitle": "LCOF2",
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"boundTopicId": 910301,
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"title": "向下的路径节点之和",
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"titleSlug": "6eUYwP",
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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": "向下的路径节点之和",
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"translatedContent": "<p>给定一个二叉树的根节点 <code>root</code> ，和一个整数 <code>targetSum</code> ，求该二叉树里节点值之和等于 <code>targetSum</code> 的 <strong>路径</strong> 的数目。</p>\n\n<p><strong>路径</strong> 不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img src=\"https://assets.leetcode.com/uploads/2021/04/09/pathsum3-1-tree.jpg\" style=\"width: 452px; \" /></p>\n\n<pre>\n<strong>输入：</strong>root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8\n<strong>输出：</strong>3\n<strong>解释：</strong>和等于 8 的路径有 3 条，如图所示。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22\n<strong>输出：</strong>3\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li>二叉树的节点个数的范围是 <code>[0,1000]</code></li>\n\t<li><meta charset=\"UTF-8\" /><code>-10<sup><span style=\"font-size: 9.449999809265137px;\">9</span></sup> <= Node.val <= 10<sup><span style=\"font-size: 9.449999809265137px;\">9</span></sup></code> </li>\n\t<li><code>-1000 <= targetSum <= 1000</code> </li>\n</ul>\n\n<p> </p>\n\n<p><meta charset=\"UTF-8\" />注意：本题与主站 437 题相同：<a href=\"https://leetcode-cn.com/problems/path-sum-iii/\">https://leetcode-cn.com/problems/path-sum-iii/</a></p>\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 * @param {number} targetSum\n * @return {number}\n */\nvar pathSum = function(root, targetSum) {\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# @param {Integer} target_sum\n# @return {Integer}\ndef path_sum(root, target_sum)\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 pathSum(_ root: TreeNode?, _ targetSum: Int) -> Int {\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 pathSum(root *TreeNode, targetSum int) int {\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 pathSum(root: TreeNode?, targetSum: Int): Int {\n\n }\n}",
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