mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
180 lines
25 KiB
JSON
180 lines
25 KiB
JSON
{
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"questionId": "222",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1238,
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"title": "Count Complete Tree Nodes",
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"content": "<p>Given the <code>root</code> of a <strong>complete</strong> binary tree, return the number of the nodes in the tree.</p>\n\n<p>According to <strong><a href=\"http://en.wikipedia.org/wiki/Binary_tree#Types_of_binary_trees\" target=\"_blank\">Wikipedia</a></strong>, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between <code>1</code> and <code>2<sup>h</sup></code> nodes inclusive at the last level <code>h</code>.</p>\n\n<p>Design an algorithm that runs in less than <code data-stringify-type=\"code\">O(n)</code> time complexity.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/01/14/complete.jpg\" style=\"width: 372px; height: 302px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3,4,5,6]\n<strong>Output:</strong> 6\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = []\n<strong>Output:</strong> 0\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [1]\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>[0, 5 * 10<sup>4</sup>]</code>.</li>\n\t<li><code>0 <= Node.val <= 5 * 10<sup>4</sup></code></li>\n\t<li>The tree is guaranteed to be <strong>complete</strong>.</li>\n</ul>\n",
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"translatedTitle": "完全二叉树的节点个数",
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"translatedContent": "<p>给你一棵<strong> 完全二叉树</strong> 的根节点 <code>root</code> ，求出该树的节点个数。</p>\n\n<p><a href=\"https://baike.baidu.com/item/%E5%AE%8C%E5%85%A8%E4%BA%8C%E5%8F%89%E6%A0%91/7773232?fr=aladdin\">完全二叉树</a> 的定义如下：在完全二叉树中，除了最底层节点可能没填满外，其余每层节点数都达到最大值，并且最下面一层的节点都集中在该层最左边的若干位置。若最底层为第 <code>h</code> 层，则该层包含 <code>1~ 2<sup>h</sup></code> 个节点。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/01/14/complete.jpg\" style=\"width: 372px; height: 302px;\" />\n<pre>\n<strong>输入：</strong>root = [1,2,3,4,5,6]\n<strong>输出：</strong>6\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = []\n<strong>输出：</strong>0\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [1]\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>树中节点的数目范围是<code>[0, 5 * 10<sup>4</sup>]</code></li>\n\t<li><code>0 <= Node.val <= 5 * 10<sup>4</sup></code></li>\n\t<li>题目数据保证输入的树是 <strong>完全二叉树</strong></li>\n</ul>\n\n<p> </p>\n\n<p><strong>进阶：</strong>遍历树来统计节点是一种时间复杂度为 <code>O(n)</code> 的简单解决方案。你可以设计一个更快的算法吗？</p>\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 count_nodes(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 countNodes(_ root: TreeNode?) -> Int {\n\n }\n}",
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