Introduction

In this blog post, we will explore how to solve the LeetCode problem "1145" 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

Two players play a turn based game on a binary tree. We are given the root of this binary tree, and the number of nodes n in the tree. n is odd, and each node has a distinct value from 1 to n. Initially, the first player names a value x with 1 <= x <= n, and the second player names a value y with 1 <= y <= n and y != x. The first player colors the node with value x red, and the second player colors the node with value y blue. Then, the players take turns starting with the first player. In each turn, that player chooses a node of their color (red if player 1, blue if player 2) and colors an uncolored neighbor of the chosen node (either the left child, right child, or parent of the chosen node.) If (and only if) a player cannot choose such a node in this way, they must pass their turn. If both players pass their turn, the game ends, and the winner is the player that colored more nodes. You are the second player. If it is possible to choose such a y to ensure you win the game, return true. If it is not possible, return false. Example 1: Input: root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3 Output: true Explanation: The second player can choose the node with value 2. Example 2: Input: root = [1,2,3], n = 3, x = 1 Output: false Constraints: The number of nodes in the tree is n. 1 <= x <= n <= 100 n is odd. 1 <= Node.val <= n All the values of the tree are unique.

Explanation

Here's the breakdown of the solution:

• Identify the target node: First, find the node with the value x chosen by the first player.
• Calculate subtree sizes: Determine the sizes of the left subtree, right subtree, and the number of nodes above (parent side) the chosen node x.

• Check winning condition: To win, the second player needs to color more nodes than the first. This can be achieved by selecting y such that one of the components (left subtree, right subtree, or parent side) of the chosen node x has more nodes than the combined nodes of x and the other two components.

• Time Complexity: O(N), where N is the number of nodes in the tree.

• Space Complexity: O(H), where H is the height of the tree (worst case O(N), average case O(logN) for a balanced tree) due to recursion stack.

Code

     class TreeNode:
     def __init__(self, val=0, left=None, right=None):
         self.val = val
         self.left = left
         self.right = right


 class Solution:
     def btreeGameWinningMove(self, root: TreeNode, n: int, x: int) -> bool:
         self.left_count = 0
         self.right_count = 0
         self.x = x


         def count_nodes(node: TreeNode) -> int:
             if not node:
                 return 0
             left = count_nodes(node.left)
             right = count_nodes(node.right)
             if node.val == self.x:
                 self.left_count = left
                 self.right_count = right
             return 1 + left + right


         count_nodes(root)


         parent_count = n - 1 - self.left_count - self.right_count


         return self.left_count > n // 2 or self.right_count > n // 2 or parent_count > n // 2