Given two binary trees original and cloned and given a reference to a node target in the original tree.
The cloned tree is a copy of the original tree.
Return a reference to the same node in the cloned tree.
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.
Follow up: Solve the problem if repeated values on the tree are allowed.
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
treeis in the range[1, 10^4]. - The values of the nodes of the
treeare unique. targetnode is a node from theoriginaltree and is notnull.
Solution: Recursion
Traverse both trees in the same order, if original == target, return cloned.
Time complexity: O(n)
Space complexity: O(h)
C++
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// Author: Huahua class Solution { public: TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) { if (!original) return nullptr; if (original == target) return cloned; auto l = getTargetCopy(original->left, cloned->left, target); auto r = getTargetCopy(original->right, cloned->right, target); return l ? l : r; } }; |
Python3
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# Author: Huahua class Solution: def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode: if not original: return None if original == target: return cloned return self.getTargetCopy(original.left, cloned.left, target) or \ self.getTargetCopy(original.right, cloned.right, target) |
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