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Given two binary trees original and cloned, and given a reference to a node target in the original tree, the task is to find the corresponding node in the cloned tree.

Both trees are identical in structure and the nodes contain the same values, only their memory locations are different. The problem guarantees that the target node is a node in the original tree and we need to return the corresponding node in the cloned tree.

Clarifying Questions

1. Are the values in the nodes unique?
   • Yes, but we’ll be utilizing the structure and reference of the target node, not just the value.
2. Can there be any additional constraints like maximum height of the tree or maximum number of nodes?
   • Assume typical constraints for binary trees in competitive programming.
3. Is the tree a full, complete, or balanced binary tree?
   • The problem does not specify, so assume it can be any binary tree.
4. Can we assume that the cloned tree is always a valid clone of the original tree?
   • Yes, we can assume that cloned is a true clone of the original tree.

Strategy

The approach involves traversing both the original and cloned trees simultaneously. When the target node is found in the original tree, the node at the same position in the cloned tree will be our result.

We’ll use a simple DFS (Depth First Search) or BFS (Breadth First Search) to traverse both trees simultaneously.

Code

Here’s a Python implementation using DFS recursively:

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def getTargetCopy(self, original: TreeNode, cloned: TreeNode, target: TreeNode) -> TreeNode:
        # Base case: if the original tree node is None, return None
        if original is None:
            return None

        # If the original node is the target node, return the corresponding node in the cloned tree
        if original == target:
            return cloned

        # Search in the left subtree
        left_result = self.getTargetCopy(original.left, cloned.left, target)
        if left_result is not None:
            return left_result
        
        # If the target is not found in the left subtree, search in the right subtree
        return self.getTargetCopy(original.right, cloned.right, target)

Time Complexity

• Time Complexity: (O(N)), where (N) is the number of nodes in the original (and cloned) tree. This is because in the worst case, we might need to visit all nodes to find the target.

• Space Complexity: (O(H)), where (H) is the height of the tree, accounting for the recursion stack. In the worst case (an unbalanced tree), (H) can be (N), making the space complexity (O(N)). For a balanced tree, it would be (O(\log N)).

This solution efficiently finds the corresponding node by recursively traversing both trees in sync until the target node is found.

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