mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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189 lines
27 KiB
JSON
189 lines
27 KiB
JSON
{
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"questionId": "1568",
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"questionFrontendId": "1457",
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"categoryTitle": "Algorithms",
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"boundTopicId": 255889,
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"title": "Pseudo-Palindromic Paths in a Binary Tree",
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"titleSlug": "pseudo-palindromic-paths-in-a-binary-tree",
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"content": "<p>Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be <strong>pseudo-palindromic</strong> if at least one permutation of the node values in the path is a palindrome.</p>\n\n<p><em>Return the number of <strong>pseudo-palindromic</strong> paths going from the root node to leaf nodes.</em></p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/05/06/palindromic_paths_1.png\" style=\"width: 300px; height: 201px;\" /></p>\n\n<pre>\n<strong>Input:</strong> root = [2,3,1,3,1,null,1]\n<strong>Output:</strong> 2 \n<strong>Explanation:</strong> The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the red path [2,3,3], the green path [2,1,1], and the path [2,3,1]. Among these paths only red path and green path are pseudo-palindromic paths since the red path [2,3,3] can be rearranged in [3,2,3] (palindrome) and the green path [2,1,1] can be rearranged in [1,2,1] (palindrome).\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/05/07/palindromic_paths_2.png\" style=\"width: 300px; height: 314px;\" /></strong></p>\n\n<pre>\n<strong>Input:</strong> root = [2,1,1,1,3,null,null,null,null,null,1]\n<strong>Output:</strong> 1 \n<strong>Explanation:</strong> The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the green path [2,1,1], the path [2,1,3,1], and the path [2,1]. Among these paths only the green path is pseudo-palindromic since [2,1,1] can be rearranged in [1,2,1] (palindrome).\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [9]\n<strong>Output:</strong> 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 in the range <code>[1, 10<sup>5</sup>]</code>.</li>\n\t<li><code>1 <= Node.val <= 9</code></li>\n</ul>\n",
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"translatedTitle": "二叉树中的伪回文路径",
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"translatedContent": "<p>给你一棵二叉树，每个节点的值为 1 到 9 。我们称二叉树中的一条路径是 「<strong>伪回文</strong>」的，当它满足：路径经过的所有节点值的排列中，存在一个回文序列。</p>\n\n<p>请你返回从根到叶子节点的所有路径中 <strong>伪回文 </strong>路径的数目。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/05/23/palindromic_paths_1.png\" style=\"height: 201px; width: 300px;\" /></p>\n\n<pre>\n<strong>输入：</strong>root = [2,3,1,3,1,null,1]\n<strong>输出：</strong>2 \n<strong>解释：</strong>上图为给定的二叉树。总共有 3 条从根到叶子的路径：红色路径 [2,3,3] ，绿色路径 [2,1,1] 和路径 [2,3,1] 。\n 在这些路径中，只有红色和绿色的路径是伪回文路径，因为红色路径 [2,3,3] 存在回文排列 [3,2,3] ，绿色路径 [2,1,1] 存在回文排列 [1,2,1] 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/05/23/palindromic_paths_2.png\" style=\"height: 314px; width: 300px;\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>root = [2,1,1,1,3,null,null,null,null,null,1]\n<strong>输出：</strong>1 \n<strong>解释：</strong>上图为给定二叉树。总共有 3 条从根到叶子的路径：绿色路径 [2,1,1] ，路径 [2,1,3,1] 和路径 [2,1] 。\n 这些路径中只有绿色路径是伪回文路径，因为 [2,1,1] 存在回文排列 [1,2,1] 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [9]\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>给定二叉树的节点数目在范围 <code>[1, 10<sup>5</sup>]</code> 内</li>\n\t<li><code>1 <= Node.val <= 9</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 PseudoPalindromicPaths (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 {number}\n */\nvar pseudoPalindromicPaths = function(root) {\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 {Integer}\ndef pseudo_palindromic_paths (root)\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 pseudoPalindromicPaths (_ root: TreeNode?) -> 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 pseudoPalindromicPaths (root: TreeNode?): 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 pseudo_palindromic_paths (root: Option<Rc<RefCell<TreeNode>>>) -> i32 {\n\n }\n}",
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"Note that the node values of a path form a palindrome if at most one digit has an odd frequency (parity).",
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"Use a Depth First Search (DFS) keeping the frequency (parity) of the digits. Once you are in a leaf node check if at most one digit has an odd frequency (parity)."
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