Introduction

In this blog post, we will explore how to solve the LeetCode problem "971" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given the root of a binary tree with n nodes, where each node is uniquely assigned a value from 1 to n. You are also given a sequence of n values voyage, which is the desired pre-order traversal of the binary tree. Any node in the binary tree can be flipped by swapping its left and right subtrees. For example, flipping node 1 will have the following effect: Flip the smallest number of nodes so that the pre-order traversal of the tree matches voyage. Return a list of the values of all flipped nodes. You may return the answer in any order. If it is impossible to flip the nodes in the tree to make the pre-order traversal match voyage, return the list [-1]. Example 1: Input: root = [1,2], voyage = [2,1] Output: [-1] Explanation: It is impossible to flip the nodes such that the pre-order traversal matches voyage. Example 2: Input: root = [1,2,3], voyage = [1,3,2] Output: [1] Explanation: Flipping node 1 swaps nodes 2 and 3, so the pre-order traversal matches voyage. Example 3: Input: root = [1,2,3], voyage = [1,2,3] Output: [] Explanation: The tree's pre-order traversal already matches voyage, so no nodes need to be flipped. Constraints: The number of nodes in the tree is n. n == voyage.length 1 <= n <= 100 1 <= Node.val, voyage[i] <= n All the values in the tree are unique. All the values in voyage are unique.

Explanation

Here's a breakdown of the solution:

• Pre-order Traversal and Comparison: Perform a pre-order traversal of the binary tree while simultaneously comparing the visited node values with the voyage sequence.
• Flipping Nodes: If the current node's value doesn't match the expected value in voyage, check if flipping the node (swapping left and right subtrees) makes the subsequent traversal match.

• Impossibility Detection: If a mismatch occurs and flipping the node doesn't resolve it, the traversal is impossible, and we return [-1].

• Time Complexity: O(N), where N is the number of nodes in the tree. We visit each node once.

• Space Complexity: O(H), where H is the height of the tree. This is due to the recursion stack during the pre-order traversal. In the worst case (skewed tree), H = N, and in the best case (balanced tree), H = log N. The flipped list takes at most O(N) space.

Code

    from typing import List, Optional

# 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 flipMatchVoyage(self, root: Optional[TreeNode], voyage: List[int]) -> List[int]:
        flipped = []
        voyage_index = 0

        def traverse(node: Optional[TreeNode]) -> bool:
            nonlocal voyage_index

            if not node:
                return True

            if node.val != voyage[voyage_index]:
                return False

            voyage_index += 1

            if node.left and node.left.val != voyage[voyage_index]:
                flipped.append(node.val)
                node.left, node.right = node.right, node.left

            if node.left is None and node.right is not None and node.right.val != voyage[voyage_index]:
                return False

            if node.left is None and node.right is None:
                return True

            return traverse(node.left) and traverse(node.right)

        if traverse(root):
            return flipped
        else:
            return [-1]