1145. Binary Tree Coloring Game LeetCode Solution

In this guide, you will get 1145. Binary Tree Coloring Game LeetCode Solution with the best time and space complexity. The solution to Binary Tree Coloring Game problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Binary Tree Coloring Game solution in C++
Binary Tree Coloring Game solution in Java
Binary Tree Coloring Game solution in Python
Additional Resources

1145. Binary Tree Coloring Game LeetCode Solution image

Problem Statement of Binary Tree Coloring Game

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.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(h)

1145. Binary Tree Coloring Game LeetCode Solution in C++

class Solution {
 public:
  bool btreeGameWinningMove(TreeNode* root, int n, int x) {
    count(root, x);
    return max({leftCount, rightCount, n - leftCount - rightCount - 1}) > n / 2;
  }

 private:
  int leftCount;   // the number of nodes of n's left
  int rightCount;  // the number of nodes of n's right

  int count(TreeNode* root, int x) {
    if (root == nullptr)
      return 0;
    const int l = count(root->left, x);
    const int r = count(root->right, x);
    if (root->val == x) {
      leftCount = l;
      rightCount = r;
    }
    return 1 + l + r;
  }
};
/* code provided by PROGIEZ */

1145. Binary Tree Coloring Game LeetCode Solution in Java

class Solution {
  public boolean btreeGameWinningMove(TreeNode root, int n, int x) {
    count(root, x);
    return Math.max(Math.max(leftCount, rightCount), n - leftCount - rightCount - 1) > n / 2;
  }

  private int leftCount;  // the number of nodes of n's left
  private int rightCount; // the number of nodes of n's right

  private int count(TreeNode root, int x) {
    if (root == null)
      return 0;
    final int l = count(root.left, x);
    final int r = count(root.right, x);
    if (root.val == x) {
      leftCount = l;
      rightCount = r;
    }
    return 1 + l + r;
  }
}
// code provided by PROGIEZ