mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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178 lines
26 KiB
JSON
178 lines
26 KiB
JSON
{
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"questionId": "1653",
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"questionFrontendId": "1530",
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"categoryTitle": "Algorithms",
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"boundTopicId": 343649,
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"title": "Number of Good Leaf Nodes Pairs",
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"titleSlug": "number-of-good-leaf-nodes-pairs",
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"content": "<p>You are given the <code>root</code> of a binary tree and an integer <code>distance</code>. A pair of two different <strong>leaf</strong> nodes of a binary tree is said to be good if the length of <strong>the shortest path</strong> between them is less than or equal to <code>distance</code>.</p>\n\n<p>Return <em>the number of good leaf node pairs</em> in the tree.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/07/09/e1.jpg\" style=\"width: 250px; height: 250px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3,null,4], distance = 3\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> The leaf nodes of the tree are 3 and 4 and the length of the shortest path between them is 3. This is the only good pair.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/07/09/e2.jpg\" style=\"width: 250px; height: 182px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3,4,5,6,7], distance = 3\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> The good pairs are [4,5] and [6,7] with shortest path = 2. The pair [4,6] is not good because the length of ther shortest path between them is 4.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> The only good pair is [2,5].\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li>The number of nodes in the <code>tree</code> is in the range <code>[1, 2<sup>10</sup>].</code></li>\n\t<li><code>1 <= Node.val <= 100</code></li>\n\t<li><code>1 <= distance <= 10</code></li>\n</ul>\n",
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"translatedTitle": "好叶子节点对的数量",
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"translatedContent": "<p>给你二叉树的根节点 <code>root</code> 和一个整数 <code>distance</code> 。</p>\n\n<p>如果二叉树中两个 <strong>叶</strong> 节点之间的 <strong>最短路径长度</strong> 小于或者等于 <code>distance</code> ，那它们就可以构成一组 <strong>好叶子节点对</strong> 。</p>\n\n<p>返回树中 <strong>好叶子节点对的数量</strong> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p> </p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/07/26/e1.jpg\" style=\"height: 321px; width: 321px;\"></p>\n\n<pre><strong>输入：</strong>root = [1,2,3,null,4], distance = 3\n<strong>输出：</strong>1\n<strong>解释：</strong>树的叶节点是 3 和 4 ，它们之间的最短路径的长度是 3 。这是唯一的好叶子节点对。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/07/26/e2.jpg\" style=\"height: 321px; width: 441px;\"></p>\n\n<pre><strong>输入：</strong>root = [1,2,3,4,5,6,7], distance = 3\n<strong>输出：</strong>2\n<strong>解释：</strong>好叶子节点对为 [4,5] 和 [6,7] ，最短路径长度都是 2 。但是叶子节点对 [4,6] 不满足要求，因为它们之间的最短路径长度为 4 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3\n<strong>输出：</strong>1\n<strong>解释：</strong>唯一的好叶子节点对是 [2,5] 。\n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre><strong>输入：</strong>root = [100], distance = 1\n<strong>输出：</strong>0\n</pre>\n\n<p><strong>示例 5：</strong></p>\n\n<pre><strong>输入：</strong>root = [1,1,1], distance = 2\n<strong>输出：</strong>1\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>tree</code> 的节点数在 <code>[1, 2^10]</code> 范围内。</li>\n\t<li>每个节点的值都在 <code>[1, 100]</code> 之间。</li>\n\t<li><code>1 <= distance <= 10</code></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 int CountPairs(TreeNode root, int distance) {\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} distance\n * @return {number}\n */\nvar countPairs = function(root, distance) {\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} distance\n# @return {Integer}\ndef count_pairs(root, distance)\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 countPairs(_ root: TreeNode?, _ distance: 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 countPairs(root *TreeNode, distance int) int {\n\n}",
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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 countPairs(root: TreeNode, distance: Int): Int = {\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 countPairs(root: TreeNode?, distance: Int): Int {\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 count_pairs(root: Option<Rc<RefCell<TreeNode>>>, distance: i32) -> i32 {\n\n }\n}",
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"Start DFS from each leaf node. stop the DFS when the number of steps done > distance.",
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"If you reach another leaf node within distance steps, add 1 to the answer.",
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"Note that all pairs will be counted twice so divide the answer by 2."
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