{ "data": { "question": { "questionId": "617", "questionFrontendId": "617", "categoryTitle": "Algorithms", "boundTopicId": 1230, "title": "Merge Two Binary Trees", "titleSlug": "merge-two-binary-trees", "content": "

You are given two binary trees root1 and root2.

\n\n

Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree.

\n\n

Return the merged tree.

\n\n

Note: The merging process must start from the root nodes of both trees.

\n\n

 

\n

Example 1:

\n\"\"\n
\nInput: root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]\nOutput: [3,4,5,5,4,null,7]\n
\n\n

Example 2:

\n\n
\nInput: root1 = [1], root2 = [1,2]\nOutput: [2,2]\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "合并二叉树", "translatedContent": "

给你两棵二叉树： root1 和 root2 。

\n\n

想象一下，当你将其中一棵覆盖到另一棵之上时，两棵树上的一些节点将会重叠（而另一些不会）。你需要将这两棵树合并成一棵新二叉树。合并的规则是：如果两个节点重叠，那么将这两个节点的值相加作为合并后节点的新值；否则，不为 null 的节点将直接作为新二叉树的节点。

\n\n

返回合并后的二叉树。

\n\n

注意: 合并过程必须从两个树的根节点开始。

\n\n

 

\n\n

示例 1：

\n\"\"\n
\n输入：root1 = [1,3,2,5], root2 = [2,1,3,null,4,null,7]\n输出：[3,4,5,5,4,null,7]\n
\n\n

示例 2：

\n\n
\n输入：root1 = [1], root2 = [1,2]\n输出：[2,2]\n
\n\n

 

\n\n

提示：

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0 : val)\n * this.left = (left===undefined ? null : left)\n * this.right = (right===undefined ? null : right)\n * }\n */\n/**\n * @param {TreeNode} root1\n * @param {TreeNode} root2\n * @return {TreeNode}\n */\nvar mergeTrees = function(root1, root2) {\n\n};", "__typename": "CodeSnippetNode" }, { "lang": "TypeScript", "langSlug": "typescript", "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\nfunction mergeTrees(root1: TreeNode | null, root2: TreeNode | null): TreeNode | null {\n \n};", "__typename": "CodeSnippetNode" }, { "lang": "PHP", "langSlug": "php", "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 $root1\n * @param TreeNode $root2\n * @return TreeNode\n */\n function mergeTrees($root1, $root2) {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Swift", "langSlug": "swift", "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 mergeTrees(_ root1: TreeNode?, _ root2: TreeNode?) -> TreeNode? {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Kotlin", "langSlug": "kotlin", "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 mergeTrees(root1: TreeNode?, root2: TreeNode?): TreeNode? {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Dart", "langSlug": "dart", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode {\n * int val;\n * TreeNode? left;\n * TreeNode? right;\n * TreeNode([this.val = 0, this.left, this.right]);\n * }\n */\nclass Solution {\n TreeNode? mergeTrees(TreeNode? root1, TreeNode? root2) {\n \n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Go", "langSlug": "golang", "code": "/**\n * Definition for a binary tree node.\n * type TreeNode struct {\n * Val int\n * Left *TreeNode\n * Right *TreeNode\n * }\n */\nfunc mergeTrees(root1 *TreeNode, root2 *TreeNode) *TreeNode {\n\n}", "__typename": "CodeSnippetNode" }, { "lang": "Ruby", "langSlug": "ruby", "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} root1\n# @param {TreeNode} root2\n# @return {TreeNode}\ndef merge_trees(root1, root2)\n\nend", "__typename": "CodeSnippetNode" }, { "lang": "Scala", "langSlug": "scala", "code": "/**\n * Definition for a binary tree node.\n * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {\n * var value: Int = _value\n * var left: TreeNode = _left\n * var right: TreeNode = _right\n * }\n */\nobject Solution {\n def mergeTrees(root1: TreeNode, root2: TreeNode): TreeNode = {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Rust", "langSlug": "rust", "code": "// Definition for a binary tree node.\n// #[derive(Debug, PartialEq, Eq)]\n// pub struct TreeNode {\n// pub val: i32,\n// pub left: Option>>,\n// pub right: Option>>,\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 merge_trees(root1: Option>>, root2: Option>>) -> Option>> {\n\n }\n}", "__typename": "CodeSnippetNode" }, { "lang": "Racket", "langSlug": "racket", "code": "; Definition for a binary tree node.\n#|\n\n; val : integer?\n; left : (or/c tree-node? #f)\n; right : (or/c tree-node? #f)\n(struct tree-node\n (val left right) #:mutable #:transparent)\n\n; constructor\n(define (make-tree-node [val 0])\n (tree-node val #f #f))\n\n|#\n\n(define/contract (merge-trees root1 root2)\n (-> (or/c tree-node? #f) (or/c tree-node? #f) (or/c tree-node? #f))\n )", "__typename": "CodeSnippetNode" }, { "lang": "Erlang", "langSlug": "erlang", "code": "%% Definition for a binary tree node.\n%%\n%% -record(tree_node, {val = 0 :: integer(),\n%% left = null :: 'null' | #tree_node{},\n%% right = null :: 'null' | #tree_node{}}).\n\n-spec merge_trees(Root1 :: #tree_node{} | null, Root2 :: #tree_node{} | null) -> #tree_node{} | null.\nmerge_trees(Root1, Root2) ->\n .", "__typename": "CodeSnippetNode" }, { "lang": "Elixir", "langSlug": "elixir", "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 merge_trees(root1 :: TreeNode.t | nil, root2 :: TreeNode.t | nil) :: TreeNode.t | nil\n def merge_trees(root1, root2) do\n \n end\nend", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"460.7K\", \"totalSubmission\": \"580.9K\", \"totalAcceptedRaw\": 460684, \"totalSubmissionRaw\": 580885, \"acRate\": \"79.3%\"}", "hints": [], "solution": null, "status": null, "sampleTestCase": "[1,3,2,5]\n[2,1,3,null,4,null,7]", "metaData": "{\n \"name\": \"mergeTrees\",\n \"params\": [\n {\n \"name\": \"root1\",\n \"type\": \"TreeNode\",\n \"dealloc\": false\n },\n {\n \"name\": \"root2\",\n \"type\": \"TreeNode\",\n \"dealloc\": false\n }\n ],\n \"return\": {\n \"type\": \"TreeNode\"\n }\n}", "judgerAvailable": true, "judgeType": "large", "mysqlSchemas": [], "enableRunCode": true, "envInfo": "{\"cpp\":[\"C++\",\"

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