Q (Leetcode 572): Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values as a subtree of s. A subtree of s is a tree consists of a node in s and all of this node’s descendants. The tree s could also be considered as a subtree of itself.

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https://leetcode.com/problems/subtree-of-another-tree/solution/ relies (as “primary check”) on the payloads of each tree node, but in some trees, all payloads are empty or all payloads are either True or False. In these cases, the comparison of payload is only usable as a secondary check. The primary check must be structural. See key in a tree node

The O(N+K) is plain wrong.

I guess the 2^{nd} solution (and possibly 1^{st} solution) would compare every node in S to the root of T. I think there are more efficient solutions using subtree size and subtree height as secondary checks – more reliable than payload check.

My solution below uses BFT + pre/post/in-order walk !

— Preliminary step: post-order walk to get subtree-size, subtree-height at each S node + T root. (I will skip other T nodes.). Suppose T is size 22 height 4. We will look for any node Y of size 22 and height 4 and a matching payload. This would eliminate lots of S nodes:

If T height is more than 2, then lots of low-level S nodes are eliminated.

If T height is 2 or 1, then T size would be at most 3. Most high-level S nodes are eliminated.

— Solution 1: For both T and S, We take in-order walk to assign incremental IDs, then take pre-order walk to produce an array of IDs that represent the tree structure.

Can We run a level-aware BST. Only one level need to be examined … wrong!

I think the in-order walk itself is what I need. Find any node Y in S that matches size+height+payload of T root. Suppose ID(Y)=44 but ID(T root) = 4, then simply shift down by 40 and do a linear scan. Must check payload, but not height/size.

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