{ "data": { "question": { "questionId": "1000009", "questionFrontendId": "面试题 04.12", "categoryTitle": "LCCI", "boundTopicId": 91459, "title": "Paths with Sum LCCI", "titleSlug": "paths-with-sum-lcci", "content": "

You are given a binary tree in which each node contains an integer value (which might be positive or negative). Design an algorithm to count the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

\r\n\r\n

Example:
\r\nGiven the following tree and  sum = 22,

\r\n\r\n
\r\n              5\r\n             / \\\r\n            4   8\r\n           /   / \\\r\n          11  13  4\r\n         /  \\    / \\\r\n        7    2  5   1\r\n
\r\n\r\n

Output:

\r\n\r\n
\r\n3\r\nExplanation: Paths that have sum 22 are: [5,4,11,2], [5,8,4,5], [4,11,7]
\r\n\r\n

Note:

\r\n\r\n\r\n", "translatedTitle": "求和路径", "translatedContent": "

给定一棵二叉树,其中每个节点都含有一个整数数值(该值或正或负)。设计一个算法,打印节点数值总和等于某个给定值的所有路径的数量。注意,路径不一定非得从二叉树的根节点或叶节点开始或结束,但是其方向必须向下(只能从父节点指向子节点方向)。

\n\n

示例:
\n给定如下二叉树,以及目标和 sum = 22

\n\n
              5\n             / \\\n            4   8\n           /   / \\\n          11  13  4\n         /  \\    / \\\n        7    2  5   1\n
\n\n

返回:

\n\n
3\n解释:和为 22 的路径有:[5,4,11,2], [5,8,4,5], [4,11,7]
\n\n

提示:

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0 : val)\n * this.left = (left===undefined ? null : left)\n * this.right = (right===undefined ? null : right)\n * }\n * }\n */\n\nfunction pathSum(root: TreeNode | null, sum: number): number {\n\n};", "__typename": "CodeSnippetNode" }, { "lang": "PHP", "langSlug": "php", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * public $val = null;\n * public $left = null;\n * public $right = null;\n * function __construct($value) { $this->val = $value; }\n * }\n */\nclass Solution {\n\n /**\n * @param TreeNode $root\n * @param Integer $sum\n * @return Integer\n */\n function pathSum($root, $sum) {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Swift", "langSlug": "swift", "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 */\nclass Solution {\n func pathSum(_ root: TreeNode?, _ sum: Int) -> Int {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Kotlin", "langSlug": "kotlin", "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?, sum: Int): Int {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Dart", "langSlug": "dart", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * int val;\n * TreeNode? left;\n * TreeNode? right;\n * TreeNode([this.val = 0, this.left, this.right]);\n * }\n */\nclass Solution {\n int pathSum(TreeNode? root, int sum) {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Go", "langSlug": "golang", "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, sum int) int {\n\n}", "__typename": "CodeSnippetNode" }, { "lang": "Ruby", "langSlug": "ruby", "code": "# Definition for a binary tree node.\n# class TreeNode\n# attr_accessor :val, :left, :right\n# def initialize(val)\n# @val = val\n# @left, @right = nil, nil\n# end\n# end\n\n# @param {TreeNode} root\n# @param {Integer} sum\n# @return {Integer}\ndef path_sum(root, sum)\n\nend", "__typename": "CodeSnippetNode" }, { "lang": "Scala", "langSlug": "scala", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode(var _value: Int) {\n * var value: Int = _value\n * var left: TreeNode = null\n * var right: TreeNode = null\n * }\n */\nobject Solution {\n def pathSum(root: TreeNode, sum: Int): Int = {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Rust", "langSlug": "rust", "code": "// Definition for a binary tree node.\n// #[derive(Debug, PartialEq, Eq)]\n// pub struct TreeNode {\n// pub val: i32,\n// pub left: Option>>,\n// pub right: Option>>,\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 path_sum(root: Option>>, sum: i32) -> i32 {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Racket", "langSlug": "racket", "code": "; Definition for a binary tree node.\n#|\n\n; val : integer?\n; left : (or/c tree-node? #f)\n; right : (or/c tree-node? #f)\n(struct tree-node\n (val left right) #:mutable #:transparent)\n\n; constructor\n(define (make-tree-node [val 0])\n (tree-node val #f #f))\n\n|#\n\n(define/contract (path-sum root sum)\n (-> (or/c tree-node? #f) exact-integer? exact-integer?)\n\n )", "__typename": "CodeSnippetNode" }, { "lang": "Erlang", "langSlug": "erlang", "code": "%% Definition for a binary tree node.\n%%\n%% -record(tree_node, {val = 0 :: integer(),\n%% left = null :: 'null' | #tree_node{},\n%% right = null :: 'null' | #tree_node{}}).\n\n-spec path_sum(Root :: #tree_node{} | null, Sum :: integer()) -> integer().\npath_sum(Root, Sum) ->\n .", "__typename": "CodeSnippetNode" }, { "lang": "Elixir", "langSlug": "elixir", "code": "# Definition for a binary tree node.\n#\n# defmodule TreeNode do\n# @type t :: %__MODULE__{\n# val: integer,\n# left: TreeNode.t() | nil,\n# right: TreeNode.t() | nil\n# }\n# defstruct val: 0, left: nil, right: nil\n# end\n\ndefmodule Solution do\n @spec path_sum(root :: TreeNode.t | nil, sum :: integer) :: integer\n def path_sum(root, sum) do\n\n end\nend", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"31.1K\", \"totalSubmission\": \"63.4K\", \"totalAcceptedRaw\": 31096, \"totalSubmissionRaw\": 63372, \"acRate\": \"49.1%\"}", "hints": [ "尝试简化问题。如果路径必须从根开始会如何?", "不要忘记路径可能会重叠。例如,如果你正在寻找总和6,那么路径1 -> 3 -> 2和1 -> 3 -> 2 -> 4 -> 6 -> 2都是有效的。", "如果每条路径必须从根开始,就从根开始遍历所有可能的路径。可以在遍历的同时追踪和,每次找到一个路径满足我们的目标和,就增加totalpaths的值。现在,如何将它扩展到可以在任何地方开始呢?记住:只需要一个蛮力算法即可完成。你可以稍后再优化。", "为了将其扩展到从任何地方开始的路径,我们可以对所有节点重复此过程。", "如果你已经设计了以上描述的算法,那么在平衡树中你会有一个O(NlogN)的算法。这是因为共N个节点,在最坏情况下,每个节点的深度是O(logN)。节点上方的每个节点都会访问一次。因此,N个节点将被访问O(logN)的时间。有一种优化算法,其运行时间为O(N)。", "在当前的蛮力算法中重复了什么工作?", "从根开始考虑每个路径(有n个这样的路径)作为一个数组。该蛮力算法具体运作如下:拿着每个数组来寻找所有具有特定和的连续子序列。我们这样做是计算了所有子数组以及它们的和。把目光聚焦在这个小问题上可能会大有裨益。给定一个数组,你如何寻找具有特定和的所有连续子序列?同样,想想蛮力算法中的重复工作。", "我们正在寻找和为targetSum的子数组。注意,可以在常数时间得到runningSumi的值,这是从元素0到元素i的和。一个从i到j的子数组和为targetSum,则 runningSumi-1 + targetSum必须等于runningSumj(试着画一个数组或一条数字线)。随着往下走,可以追踪runningSum,那么如何能快速查找i对应的使前面等式成立的值?", "尝试使用一个散列表,从runningSum的值映射到使用runningSum元素的个数。", "一旦你完成了这样的算法,找出了和为给定值的所有连续子数组,试着将它应用到一棵树上。请记住,在遍历和修改散列表时,你可能需要在遍历回来时将散列表的“损坏”逆转。" ], "solution": null, "status": null, "sampleTestCase": "[5,4,8,11,null,13,4,7,2,null,null,5,1]\n22", "metaData": "{\r\n \"name\": \"pathSum\",\r\n \"params\": [\r\n {\r\n \"name\": \"root\",\r\n \"type\": \"TreeNode\"\r\n },\r\n {\r\n \"name\": \"sum\",\r\n \"type\": \"integer\"\r\n }\r\n ],\r\n \"return\": {\r\n \"type\": \"integer\"\r\n }\r\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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