What methods can be utilized to efficiently determine if one N-ary tree is a subtree of another, besides simply comparing their pre-order strings?

When faced with a problem of determining if one binary tree is a subtree of another, the usual method involves generating strings by performing pre-order traversals on both trees and then using indexOf to check for the presence of one string in the other. However, this approach can be inefficient due to the expense involved in string generation and comparison.

To tackle this issue, I am contemplating the use of a HashMap to store node values as keys and their corresponding children as values. By comparing the structures of the two trees based on these stored values, it might be possible to determine sub-tree relationships without relying on costly string operations.

But the question remains - how exactly should this be achieved? The concept seems promising, but putting it into practice is where I'm getting stuck. Any guidance or suggestions would be greatly appreciated.

Here is an excerpt from the given code:

// Define the Node interface
interface Node {
  value: string ;
  left?: Node ;
  right?: Node ;
}
function checkSubtree(T1: Node , T2: Node ): boolean {
  return PreOrderTraversal(T1).indexOf(PreOrderTraversal(T2)) > - 1 ;
}
function stringFromPreOrder(tree: Node ): string {
  if (!tree) {
    return "" ;
}
  return tree.value + PreOrderTraversal(tree.left) + PreOrderTraversal(tree.right);

Considering the structure for n-ary trees will involve arrays for children nodes, a revised function must account for this difference while avoiding expensive string concatenation. The aim is to improve efficiency when determining sub-tree relationships for such multi-branched trees.

Answer №1

For more information on n-ary trees, you can visit this link.

Did you know that it is possible to convert n-ary trees to binary trees and vice versa?

If you are interested in finding the subtree of a binary tree, there is already an algorithm available. You can check it out here.

I'm not entirely convinced if this is the best approach, but I wanted to share the idea with you.

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