能否通过设计二叉搜索树遍历最长右路径求解最长递增子序列问题？

Great question! Let’s break down whether using a Binary Search Tree (BST) and traversing its longest right path can solve the Longest Increasing Subsequence (LIS) problem.

First: Why a Naive BST Approach Fails #

If you just insert elements into a standard BST in sequence and then look for the longest right path, this won’t work. Here’s why:

• A standard BST’s structure depends on insertion order, not the inherent increasing subsequences that skip elements. For example, take the sequence [3,1,2,4]:
  • Insert 3 → BST root is 3 (path length 1)
  • Insert 1 → left child of 3 (path length 1)
  • Insert 2 → right child of 1 (path length 2)
  • Insert 4 → right child of 3 (path length 2)
• The longest right path here is either 3→4 or 1→2 (length 2), but the actual LIS is [1,2,4] (length 3). The BST can’t connect 2 and 4 because 4 was inserted after 3, not 2.

A Modified BST Approach That Works #

You can adapt the BST to align with the greedy + binary search logic that’s commonly used for LIS. Here’s how to structure it:

• The BST will store nodes where each node’s key is the smallest possible tail value of an increasing subsequence of a specific length. The node’s value will track the length of that subsequence.
• For each element x in your input sequence:
  1. Find the predecessor: Search the BST for the largest key that’s smaller than x. Let its subsequence length be l. This means x can extend that subsequence to length l+1.
  2. Clean up redundant nodes: Delete all nodes in the BST where the key is ≥ x and the subsequence length is ≤ l+1. These nodes are useless because x is a smaller tail for the same or longer subsequences, making future extensions easier.
  3. Insert the new node: Add x to the BST with a subsequence length of l+1.
• Once all elements are processed, the longest right path in this modified BST will exactly match the length of the LIS. Each right child in this path represents a longer subsequence with a larger (but minimal possible) tail value.

Example Walkthrough for [3,1,2,4] #

1. Insert 3 → Node: (key:3, length:1) (BST root)
2. Insert 1:
   • No predecessor smaller than 1, so length = 1.
   • Delete nodes ≥1 with length ≤1 (the 3 node).
   • Insert (key:1, length:1) as root.
3. Insert 2:
   • Predecessor is 1 (length 1), so new length = 2.
   • No nodes ≥2 with length ≤2 exist.
   • Insert (key:2, length:2) as right child of 1.
4. Insert 4:
   • Predecessor is 2 (length 2), so new length =3.
   • No nodes ≥4 with length ≤3 exist.
   • Insert (key:4, length:3) as right child of 2.

• The longest right path is 1→2→4 (length 3), which matches the LIS length.

Pros and Cons of This Approach #

Pros #

• Dynamic updates: Unlike the static greedy+binary search method, this BST approach works well if you need to add elements to the sequence and update the LIS in real time.
• Time complexity: With a balanced BST (like an AVL tree or Red-Black tree), each insertion/search/delete operation takes O(log n) time, leading to an overall O(n log n) time complexity—same as the optimal LIS algorithms.

Cons #

• Implementation complexity: This is more involved than the standard greedy+binary search method. You’ll need to handle balanced BST operations (or use a built-in balanced tree structure if your language provides one, like TreeSet in Java).
• Balancing requirement: A naive unbalanced BST could degenerate into a linked list in worst-case scenarios (e.g., a fully increasing sequence), leading to O(n) time per operation instead of O(log n).

Final Verdict #

A naive BST + longest right path won’t solve the LIS problem, but a modified balanced BST that tracks the minimal tail values for each subsequence length will work. The longest right path in this tailored BST will indeed give you the length of the LIS.

内容的提问来源于stack exchange，提问作者user6161815