mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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182 lines
26 KiB
JSON
182 lines
26 KiB
JSON
{
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"questionId": "114",
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"questionFrontendId": "114",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1757,
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"title": "Flatten Binary Tree to Linked List",
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"titleSlug": "flatten-binary-tree-to-linked-list",
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"content": "<p>Given the <code>root</code> of a binary tree, flatten the tree into a "linked list":</p>\n\n<ul>\n\t<li>The "linked list" should use the same <code>TreeNode</code> class where the <code>right</code> child pointer points to the next node in the list and the <code>left</code> child pointer is always <code>null</code>.</li>\n\t<li>The "linked list" should be in the same order as a <a href=\"https://en.wikipedia.org/wiki/Tree_traversal#Pre-order,_NLR\" target=\"_blank\"><strong>pre-order</strong><strong> traversal</strong></a> of the binary tree.</li>\n</ul>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/01/14/flaten.jpg\" style=\"width: 500px; height: 226px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,5,3,4,null,6]\n<strong>Output:</strong> [1,null,2,null,3,null,4,null,5,null,6]\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = []\n<strong>Output:</strong> []\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> root = [0]\n<strong>Output:</strong> [0]\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, 2000]</code>.</li>\n\t<li><code>-100 <= Node.val <= 100</code></li>\n</ul>\n\n<p> </p>\n<strong>Follow up:</strong> Can you flatten the tree in-place (with <code>O(1)</code> extra space)?",
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"translatedTitle": "二叉树展开为链表",
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"translatedContent": "<p>给你二叉树的根结点 <code>root</code> ，请你将它展开为一个单链表：</p>\n\n<ul>\n\t<li>展开后的单链表应该同样使用 <code>TreeNode</code> ，其中 <code>right</code> 子指针指向链表中下一个结点，而左子指针始终为 <code>null</code> 。</li>\n\t<li>展开后的单链表应该与二叉树 <a href=\"https://baike.baidu.com/item/%E5%85%88%E5%BA%8F%E9%81%8D%E5%8E%86/6442839?fr=aladdin\" target=\"_blank\"><strong>先序遍历</strong></a> 顺序相同。</li>\n</ul>\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/flaten.jpg\" style=\"width: 500px; height: 226px;\" />\n<pre>\n<strong>输入：</strong>root = [1,2,5,3,4,null,6]\n<strong>输出：</strong>[1,null,2,null,3,null,4,null,5,null,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>[]\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>root = [0]\n<strong>输出：</strong>[0]\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中结点数在范围 <code>[0, 2000]</code> 内</li>\n\t<li><code>-100 <= Node.val <= 100</code></li>\n</ul>\n\n<p> </p>\n\n<p><strong>进阶：</strong>你可以使用原地算法（<code>O(1)</code> 额外空间）展开这棵树吗？</p>\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 void Flatten(TreeNode root) {\n\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\n/**\n Do not return anything, modify root in-place instead.\n */\nfunction flatten(root: TreeNode | null): void {\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 $root\n * @return NULL\n */\n function flatten($root) {\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 flatten(_ root: TreeNode?) {\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 flatten(root: TreeNode?): Unit {\n\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 {Void} Do not return anything, modify root in-place instead.\ndef flatten(root)\n\nend",
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"If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal."
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