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.
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
Constraints:
- The number of nodes in the
tree
is in the range[1, 104]
. - The values of the nodes of the
tree
are unique. -
target
node is a node from theoriginal
tree and is notnull
.
Follow up: Could you solve the problem if repeated values on the tree are allowed?
SOLUTION:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
if not p or not q:
if p or q:
return False
return True
if p.val == q.val and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right):
return True
def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode:
nodes = [(original, cloned)]
while len(nodes) > 0:
og, cl = nodes.pop()
if self.isSameTree(og, target):
return cl
if og and cl:
nodes.append((og.left, cl.left))
nodes.append((og.right, cl.right))
return None
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