mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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178 lines
28 KiB
JSON
178 lines
28 KiB
JSON
{
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"questionId": "1011",
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"questionFrontendId": "971",
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"categoryTitle": "Algorithms",
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"boundTopicId": 2763,
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"title": "Flip Binary Tree To Match Preorder Traversal",
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"titleSlug": "flip-binary-tree-to-match-preorder-traversal",
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"content": "<p>You are given the <code>root</code> of a binary tree with <code>n</code> nodes, where each node is uniquely assigned a value from <code>1</code> to <code>n</code>. You are also given a sequence of <code>n</code> values <code>voyage</code>, which is the <strong>desired</strong> <a href=\"https://en.wikipedia.org/wiki/Tree_traversal#Pre-order\" target=\"_blank\"><strong>pre-order traversal</strong></a> of the binary tree.</p>\n\n<p>Any node in the binary tree can be <strong>flipped</strong> by swapping its left and right subtrees. For example, flipping node 1 will have the following effect:</p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/02/15/fliptree.jpg\" style=\"width: 400px; height: 187px;\" />\n<p>Flip the <strong>smallest</strong> number of nodes so that the <strong>pre-order traversal</strong> of the tree <strong>matches</strong> <code>voyage</code>.</p>\n\n<p>Return <em>a list of the values of all <strong>flipped</strong> nodes. You may return the answer in <strong>any order</strong>. If it is <strong>impossible</strong> to flip the nodes in the tree to make the pre-order traversal match </em><code>voyage</code><em>, return the list </em><code>[-1]</code>.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/01/02/1219-01.png\" style=\"width: 150px; height: 205px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2], voyage = [2,1]\n<strong>Output:</strong> [-1]\n<strong>Explanation:</strong> It is impossible to flip the nodes such that the pre-order traversal matches voyage.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/01/02/1219-02.png\" style=\"width: 150px; height: 142px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3], voyage = [1,3,2]\n<strong>Output:</strong> [1]\n<strong>Explanation:</strong> Flipping node 1 swaps nodes 2 and 3, so the pre-order traversal matches voyage.</pre>\n\n<p><strong>Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/01/02/1219-02.png\" style=\"width: 150px; height: 142px;\" />\n<pre>\n<strong>Input:</strong> root = [1,2,3], voyage = [1,2,3]\n<strong>Output:</strong> []\n<strong>Explanation:</strong> The tree's pre-order traversal already matches voyage, so no nodes need to be flipped.\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>n == voyage.length</code></li>\n\t<li><code>1 <= n <= 100</code></li>\n\t<li><code>1 <= Node.val, voyage[i] <= n</code></li>\n\t<li>All the values in the tree are <strong>unique</strong>.</li>\n\t<li>All the values in <code>voyage</code> are <strong>unique</strong>.</li>\n</ul>\n",
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"translatedTitle": "翻转二叉树以匹配先序遍历",
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"translatedContent": "<p>给你一棵二叉树的根节点 <code>root</code> ，树中有 <code>n</code> 个节点，每个节点都有一个不同于其他节点且处于 <code>1</code> 到 <code>n</code> 之间的值。</p>\n\n<p>另给你一个由 <code>n</code> 个值组成的行程序列 <code>voyage</code> ，表示 <strong>预期</strong> 的二叉树 <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> 结果。</p>\n\n<p>通过交换节点的左右子树，可以 <strong>翻转</strong> 该二叉树中的任意节点。例，翻转节点 1 的效果如下：</p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/02/15/fliptree.jpg\" style=\"width: 400px; height: 187px;\" />\n<p>请翻转 <strong>最少 </strong>的树中节点，使二叉树的 <strong>先序遍历</strong> 与预期的遍历行程 <code>voyage</code> <strong>相匹配</strong> 。 </p>\n\n<p>如果可以，则返回 <strong>翻转的</strong> 所有节点的值的列表。你可以按任何顺序返回答案。如果不能，则返回列表 <code>[-1]</code>。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/01/02/1219-01.png\" style=\"width: 150px; height: 205px;\" />\n<pre>\n<strong>输入：</strong>root = [1,2], voyage = [2,1]\n<strong>输出：</strong>[-1]\n<strong>解释：</strong>翻转节点无法令先序遍历匹配预期行程。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/01/02/1219-02.png\" style=\"width: 150px; height: 142px;\" />\n<pre>\n<strong>输入：</strong>root = [1,2,3], voyage = [1,3,2]\n<strong>输出：</strong>[1]\n<strong>解释：</strong>交换节点 2 和 3 来翻转节点 1 ，先序遍历可以匹配预期行程。</pre>\n\n<p><strong>示例 3：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/01/02/1219-02.png\" style=\"width: 150px; height: 142px;\" />\n<pre>\n<strong>输入：</strong>root = [1,2,3], voyage = [1,2,3]\n<strong>输出：</strong>[]\n<strong>解释：</strong>先序遍历已经匹配预期行程，所以不需要翻转节点。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li>树中的节点数目为 <code>n</code></li>\n\t<li><code>n == voyage.length</code></li>\n\t<li><code>1 <= n <= 100</code></li>\n\t<li><code>1 <= Node.val, voyage[i] <= n</code></li>\n\t<li>树中的所有值 <strong>互不相同</strong></li>\n\t<li><code>voyage</code> 中的所有值 <strong>互不相同</strong></li>\n</ul>\n",
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