1379. Find a Corresponding Node of a Binary Tree in a Clone of That Tree

问题：

• 给定两棵二叉树。
  • 原二叉树original
  • 拷贝二叉树cloned
• 以及原二叉树上的节点target

求拷贝二叉树上同一个位置的节点。

Example 1:
Input: tree = [7,4,3,null,null,6,19], target = 3
Output: 3
Explanation: 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.

Example 2:
Input: tree = [7], target =  7
Output: 7

Example 3:
Input: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4
Output: 4

Example 4:
Input: tree = [1,2,3,4,5,6,7,8,9,10], target = 5
Output: 5

Example 5:
Input: tree = [1,2,null,3], target = 2
Output: 2
 

Constraints:
The number of nodes in the tree is in the range [1, 10^4].
The values of the nodes of the tree are unique.
target node is a node from the original tree and is not null.

example 1:

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example 2:

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example 3:

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example 4:

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example 5:

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解法：DFS

• 状态：
  • 原二叉树的当前节点 original
  • 拷贝二叉树的当前节点cloned
• 退出递归条件：
  • 当前节点original为nullptr，返回 nullptr
  • 当前节点original==target，返回 cloned
• 选择：
  • 递归左子树：original->left，cloned->left
  • 若左子树返回nullptr，继续递归右子树：original->right，cloned->right
  • 返回右子树的结果。

代码参考：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) {
        if(original == nullptr) return original;
        if(original == target) return cloned;
        TreeNode* left = getTargetCopy(original->left, cloned->left, target);
        if(!left) return getTargetCopy(original->right, cloned->right, target);
        return left;
    }
};