Description
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 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.
分析
题目的意思是：这道题给了两颗二叉树，和一个target结点，其中是二叉树是复制过来的，然后要找出对应的复制的二叉树的target结点并返回。思路也很直接，遍历复制的二叉树找到对应的结点返回就行了。好像过于简单，我有点慌…
代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode:
if(original is None):
return None
l=self.getTargetCopy(original.left,cloned.left,target)
r=self.getTargetCopy(original.right,cloned.right,target)
if(original.val==target.val):
return cloned
return l or r