mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
190 lines
30 KiB
JSON
190 lines
30 KiB
JSON
{
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"data": {
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"question": {
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"questionId": "2217",
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"questionFrontendId": "2096",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1133481,
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"title": "Step-By-Step Directions From a Binary Tree Node to Another",
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"titleSlug": "step-by-step-directions-from-a-binary-tree-node-to-another",
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"content": "<p>You are given the <code>root</code> of a <strong>binary tree</strong> with <code>n</code> nodes. Each node is uniquely assigned a value from <code>1</code> to <code>n</code>. You are also given an integer <code>startValue</code> representing the value of the start node <code>s</code>, and a different integer <code>destValue</code> representing the value of the destination node <code>t</code>.</p>\n\n<p>Find the <strong>shortest path</strong> starting from node <code>s</code> and ending at node <code>t</code>. Generate step-by-step directions of such path as a string consisting of only the <strong>uppercase</strong> letters <code>'L'</code>, <code>'R'</code>, and <code>'U'</code>. Each letter indicates a specific direction:</p>\n\n<ul>\n\t<li><code>'L'</code> means to go from a node to its <strong>left child</strong> node.</li>\n\t<li><code>'R'</code> means to go from a node to its <strong>right child</strong> node.</li>\n\t<li><code>'U'</code> means to go from a node to its <strong>parent</strong> node.</li>\n</ul>\n\n<p>Return <em>the step-by-step directions of the <strong>shortest path</strong> from node </em><code>s</code><em> to node</em> <code>t</code>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/15/eg1.png\" style=\"width: 214px; height: 163px;\" />\n<pre>\n<strong>Input:</strong> root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6\n<strong>Output:</strong> "UURL"\n<strong>Explanation:</strong> The shortest path is: 3 → 1 → 5 → 2 → 6.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/15/eg2.png\" style=\"width: 74px; height: 102px;\" />\n<pre>\n<strong>Input:</strong> root = [2,1], startValue = 2, destValue = 1\n<strong>Output:</strong> "L"\n<strong>Explanation:</strong> The shortest path is: 2 → 1.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the tree is <code>n</code>.</li>\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= Node.val <= n</code></li>\n\t<li>All the values in the tree are <strong>unique</strong>.</li>\n\t<li><code>1 <= startValue, destValue <= n</code></li>\n\t<li><code>startValue != destValue</code></li>\n</ul>\n",
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"translatedTitle": "从二叉树一个节点到另一个节点每一步的方向",
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"translatedContent": "<p>给你一棵 <strong>二叉树</strong> 的根节点 <code>root</code> ，这棵二叉树总共有 <code>n</code> 个节点。每个节点的值为 <code>1</code> 到 <code>n</code> 中的一个整数，且互不相同。给你一个整数 <code>startValue</code> ，表示起点节点 <code>s</code> 的值，和另一个不同的整数 <code>destValue</code> ，表示终点节点 <code>t</code> 的值。</p>\n\n<p>请找到从节点 <code>s</code> 到节点 <code>t</code> 的 <strong>最短路径</strong> ，并以字符串的形式返回每一步的方向。每一步用 <strong>大写</strong> 字母 <code>'L'</code> ，<code>'R'</code> 和 <code>'U'</code> 分别表示一种方向：</p>\n\n<ul>\n\t<li><code>'L'</code> 表示从一个节点前往它的 <strong>左孩子</strong> 节点。</li>\n\t<li><code>'R'</code> 表示从一个节点前往它的 <strong>右孩子</strong> 节点。</li>\n\t<li><code>'U'</code> 表示从一个节点前往它的 <strong>父</strong> 节点。</li>\n</ul>\n\n<p>请你返回从 <code>s</code> 到 <code>t</code> <strong>最短路径</strong> 每一步的方向。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/15/eg1.png\" style=\"width: 214px; height: 163px;\"></p>\n\n<pre><b>输入：</b>root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6\n<b>输出：</b>\"UURL\"\n<b>解释：</b>最短路径为：3 → 1 → 5 → 2 → 6 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/15/eg2.png\" style=\"width: 74px; height: 102px;\"></p>\n\n<pre><b>输入：</b>root = [2,1], startValue = 2, destValue = 1\n<b>输出：</b>\"L\"\n<b>解释：</b>最短路径为：2 → 1 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中节点数目为 <code>n</code> 。</li>\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= Node.val <= n</code></li>\n\t<li>树中所有节点的值 <strong>互不相同</strong> 。</li>\n\t<li><code>1 <= startValue, destValue <= n</code></li>\n\t<li><code>startValue != destValue</code></li>\n</ul>\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 string getDirections(TreeNode* root, int startValue, int destValue) {\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 String getDirections(TreeNode root, int startValue, int destValue) {\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 getDirections(self, root, startValue, destValue):\n \"\"\"\n :type root: Optional[TreeNode]\n :type startValue: int\n :type destValue: int\n :rtype: str\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 getDirections(self, root: Optional[TreeNode], startValue: int, destValue: int) -> str:",
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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\nchar * getDirections(struct TreeNode* root, int startValue, int destValue){\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 string GetDirections(TreeNode root, int startValue, int destValue) {\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 * @param {number} startValue\n * @param {number} destValue\n * @return {string}\n */\nvar getDirections = function(root, startValue, destValue) {\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 getDirections(root: TreeNode | null, startValue: number, destValue: number): string {\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 * @param Integer $startValue\n * @param Integer $destValue\n * @return String\n */\n function getDirections($root, $startValue, $destValue) {\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 getDirections(_ root: TreeNode?, _ startValue: Int, _ destValue: Int) -> String {\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 getDirections(root: TreeNode?, startValue: Int, destValue: Int): String {\n\n }\n}",
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"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 String getDirections(TreeNode? root, int startValue, int destValue) {\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 getDirections(root *TreeNode, startValue int, destValue int) string {\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} start_value\n# @param {Integer} dest_value\n# @return {String}\ndef get_directions(root, start_value, dest_value)\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 getDirections(root: TreeNode, startValue: Int, destValue: Int): String = {\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 get_directions(root: Option<Rc<RefCell<TreeNode>>>, start_value: i32, dest_value: i32) -> String {\n\n }\n}",
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"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 (get-directions root startValue destValue)\n (-> (or/c tree-node? #f) exact-integer? exact-integer? string?)\n\n )",
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"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 get_directions(Root :: #tree_node{} | null, StartValue :: integer(), DestValue :: integer()) -> unicode:unicode_binary().\nget_directions(Root, StartValue, DestValue) ->\n .",
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"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 get_directions(root :: TreeNode.t | nil, start_value :: integer, dest_value :: integer) :: String.t\n def get_directions(root, start_value, dest_value) do\n\n end\nend",
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"The shortest path between any two nodes in a tree must pass through their Lowest Common Ancestor (LCA). The path will travel upwards from node s to the LCA and then downwards from the LCA to node t.",
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"Find the path strings from root → s, and root → t. Can you use these two strings to prepare the final answer?",
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"Remove the longest common prefix of the two path strings to get the path LCA → s, and LCA → t. Each step in the path of LCA → s should be reversed as 'U'."
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"sampleTestCase": "[5,1,2,3,null,6,4]\n3\n6",
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