{ "data": { "question": { "questionId": "1498", "questionFrontendId": "1379", "categoryTitle": "Algorithms", "boundTopicId": 145880, "title": "Find a Corresponding Node of a Binary Tree in a Clone of That Tree", "titleSlug": "find-a-corresponding-node-of-a-binary-tree-in-a-clone-of-that-tree", "content": "

Given two binary trees original and cloned and given a reference to a node target in the original tree.

\n\n

The cloned tree is a copy of the original tree.

\n\n

Return a reference to the same node in the cloned tree.

\n\n

Note that you are not allowed to change any of the two trees or the target node and the answer must be a reference to a node in the cloned tree.

\n\n

 

\n

Example 1:

\n\"\"\n
\nInput: tree = [7,4,3,null,null,6,19], target = 3\nOutput: 3\nExplanation: In all examples the original and cloned trees are shown. The target node is a green node from the original tree. The answer is the yellow node from the cloned tree.\n
\n\n

Example 2:

\n\"\"\n
\nInput: tree = [7], target =  7\nOutput: 7\n
\n\n

Example 3:

\n\"\"\n
\nInput: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4\nOutput: 4\n
\n\n

 

\n

Constraints:

\n\n\n\n

 

\n

Follow up: Could you solve the problem if repeated values on the tree are allowed?

\n", "translatedTitle": "找出克隆二叉树中的相同节点", "translatedContent": "

给你两棵二叉树，原始树 original 和克隆树 cloned，以及一个位于原始树 original 中的目标节点 target。

\n\n

其中，克隆树 cloned 是原始树 original 的一个 副本 。

\n\n

请找出在树 cloned 中，与 target 相同 的节点，并返回对该节点的引用（在 C/C++ 等有指针的语言中返回 节点指针，其他语言返回节点本身）。

\n\n

 

\n\n

注意：你 不能 对两棵二叉树，以及 target 节点进行更改。只能 返回对克隆树 cloned 中已有的节点的引用。

\n\n\n\n

 

\n\n\n\n

示例 1:

\n\n

\"\"

\n\n
\n输入: tree = [7,4,3,null,null,6,19], target = 3\n输出: 3\n解释: 上图画出了树 original 和 cloned。target 节点在树 original 中，用绿色标记。答案是树 cloned 中的黄颜色的节点（其他示例类似）。
\n\n

示例 2:

\n\n

\"\"

\n\n
\n输入: tree = [7], target =  7\n输出: 7\n
\n\n

示例 3:

\n\n

\"\"

\n\n
\n输入: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4\n输出: 4\n
\n\n

 

\n\n

提示：

\n\n\n\n

 

\n\n

进阶：如果树中允许出现值相同的节点，将如何解答？

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