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

Similar Problems:

- CheatSheet: LeetCode For Code Interview
- CheatSheet: Common Code Problems & Follow-ups
- Tag: #binarytree

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution: with duplicate values

## Basic Ideas: tree traversal ## ## In original tree, find the path from root to the target ## ## Complexity: Time O(n), Space O(1) # 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: paths = [] # return whether in the path def dfs(root, target): nonlocal paths if not root: return False if root == target: paths.append(root) return True if dfs(root.left, target): paths.append(root) return True if dfs(root.right, target): paths.append(root) return True return False dfs(original, target) n = len(paths) p = cloned for i in range(n-2, -1, -1): if paths[i] == paths[i+1].left: p = p.left else: p = p.right return p

- Solution: with unique values

## https://code.dennyzhang.com/find-a-corresponding-node-of-a-binary-tree-in-a-clone-of-that-tree ## Basic Ideas: tree traversal ## ## Complexity: Time O(n), Space O(1) # 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: def dfs(root, targetVal): if not root: return None if root.val == targetVal: return root node = dfs(root.left, targetVal) if node: return node return dfs(root.right, targetVal) return dfs(cloned, target.val)