LeetCode 2005: Subtree Removal Game with Fibonacci Tree Solution

Master LeetCode problem 2005 (Subtree Removal Game with Fibonacci Tree), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2005. Subtree Removal Game with Fibonacci Tree

Hard

Math Dynamic Programming Tree Binary Tree Game Theory

Problem Explanation

Explanation

The problem involves a game played on a tree where each node has a value. The players take turns removing a subtree rooted at a node and gaining the total sum of the removed subtree. The player with the maximum score wins the game. The tree given is a Fibonacci tree, where each node has children in the order of Fibonacci numbers.

To solve this problem, we can use a recursive approach. We can calculate the maximum score that can be obtained starting from each node, considering the scores of its children. We can keep track of two values for each node: the maximum score if the current node is chosen, and the maximum score if the current node is not chosen.

At each node, the maximum score if the node is chosen will be the sum of the values of the current node and the maximum scores of the grandchildren nodes (as we cannot choose the children nodes). The maximum score if the node is not chosen will be the sum of the maximum scores of the children nodes.

We can then use dynamic programming to calculate these values bottom-up starting from the leaf nodes. Finally, we compare the maximum scores at the root node to determine the winner of the game.

Time complexity: O(n) where n is the number of nodes in the tree. Space complexity: O(n) for the dynamic programming array.

Solution Code

class Solution {
    public boolean removeGame(TreeNode root) {
        return dfs(root)[0] > 0;
    }

    private int[] dfs(TreeNode node) {
        if (node == null) {
            return new int[]{0, 0};
        }

        int[] left = dfs(node.left);
        int[] right = dfs(node.right);

        int pickNode = node.val + left[1] + right[1];
        int notPickNode = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);

        return new int[]{pickNode, notPickNode};
    }
}

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