mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
197 lines
28 KiB
JSON
197 lines
28 KiB
JSON
{
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"questionId": "572",
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"questionFrontendId": "572",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1702,
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"title": "Subtree of Another Tree",
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"titleSlug": "subtree-of-another-tree",
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"content": "<p>Given the roots of two binary trees <code>root</code> and <code>subRoot</code>, return <code>true</code> if there is a subtree of <code>root</code> with the same structure and node values of<code> subRoot</code> and <code>false</code> otherwise.</p>\n\n<p>A subtree of a binary tree <code>tree</code> is a tree that consists of a node in <code>tree</code> and all of this node's descendants. The tree <code>tree</code> could also be considered as a subtree of itself.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/04/28/subtree1-tree.jpg\" style=\"width: 532px; height: 400px;\" />\n<pre>\n<strong>Input:</strong> root = [3,4,5,1,2], subRoot = [4,1,2]\n<strong>Output:</strong> true\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/04/28/subtree2-tree.jpg\" style=\"width: 502px; height: 458px;\" />\n<pre>\n<strong>Input:</strong> root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]\n<strong>Output:</strong> false\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>root</code> tree is in the range <code>[1, 2000]</code>.</li>\n\t<li>The number of nodes in the <code>subRoot</code> tree is in the range <code>[1, 1000]</code>.</li>\n\t<li><code>-10<sup>4</sup> <= root.val <= 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>4</sup> <= subRoot.val <= 10<sup>4</sup></code></li>\n</ul>\n",
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"translatedTitle": "另一棵树的子树",
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"translatedContent": "<div class=\"original__bRMd\">\n<div>\n<p>给你两棵二叉树 <code>root</code> 和 <code>subRoot</code> 。检验 <code>root</code> 中是否包含和 <code>subRoot</code> 具有相同结构和节点值的子树。如果存在，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>\n\n<p>二叉树 <code>tree</code> 的一棵子树包括 <code>tree</code> 的某个节点和这个节点的所有后代节点。<code>tree</code> 也可以看做它自身的一棵子树。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/04/28/subtree1-tree.jpg\" style=\"width: 532px; height: 400px;\" />\n<pre>\n<strong>输入：</strong>root = [3,4,5,1,2], subRoot = [4,1,2]\n<strong>输出：</strong>true\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/04/28/subtree2-tree.jpg\" style=\"width: 502px; height: 458px;\" />\n<pre>\n<strong>输入：</strong>root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]\n<strong>输出：</strong>false\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>root</code> 树上的节点数量范围是 <code>[1, 2000]</code></li>\n\t<li><code>subRoot</code> 树上的节点数量范围是 <code>[1, 1000]</code></li>\n\t<li><code>-10<sup>4</sup> <= root.val <= 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>4</sup> <= subRoot.val <= 10<sup>4</sup></code></li>\n</ul>\n</div>\n</div>\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 isSubtree(root: TreeNode?, subRoot: TreeNode?): Boolean {\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# @param {TreeNode} sub_root\n# @return {Boolean}\ndef is_subtree(root, sub_root)\n\nend",
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"code": "# Definition for a binary tree node.\n#\n# defmodule TreeNode do\n# @type t :: %__MODULE__{\n# val: integer,\n# left: TreeNode.t() | nil,\n# right: TreeNode.t() | nil\n# }\n# defstruct val: 0, left: nil, right: nil\n# end\n\ndefmodule Solution do\n @spec is_subtree(root :: TreeNode.t | nil, sub_root :: TreeNode.t | nil) :: boolean\n def is_subtree(root, sub_root) do\n\n end\nend",
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"Which approach is better here- recursive or iterative?",
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"If recursive approach is better, can you write recursive function with its parameters?",
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"Two trees <b>s</b> and <b>t</b> are said to be identical if their root values are same and their left and right subtrees are identical. Can you write this in form of recursive formulae?",
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"Recursive formulae can be: \r\nisIdentical(s,t)= s.val==t.val AND isIdentical(s.left,t.left) AND isIdentical(s.right,t.right)"
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