1145. Binary Tree Coloring Game

Description

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.

Solutions

Solution 1: DFS

First, we use DFS to find the node where player 1’s colored point $x$ is located, denoted as $node$.

Next, we count the number of nodes in the left and right subtrees of $node$, denoted as $l$ and $r$ respectively, and the number of nodes in the direction of $node$’s parent node is $n - l - r - 1$. As long as $\max(l, r, n - l - r - 1) > \frac{n}{2}$, player 2 has a winning strategy.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the total number of nodes.

Python Code
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# 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 btreeGameWinningMove(self, root: Optional[TreeNode], n: int, x: int) -> bool:
        def dfs(root):
            if root is None or root.val == x:
                return root
            return dfs(root.left) or dfs(root.right)

        def count(root):
            if root is None:
                return 0
            return 1 + count(root.left) + count(root.right)

        node = dfs(root)
        l, r = count(node.left), count(node.right)
        return max(l, r, n - l - r - 1) > n // 2

Java Code
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean btreeGameWinningMove(TreeNode root, int n, int x) {
        TreeNode node = dfs(root, x);
        int l = count(node.left);
        int r = count(node.right);
        return Math.max(Math.max(l, r), n - l - r - 1) > n / 2;
    }

    private TreeNode dfs(TreeNode root, int x) {
        if (root == null || root.val == x) {
            return root;
        }
        TreeNode node = dfs(root.left, x);
        return node == null ? dfs(root.right, x) : node;
    }

    private int count(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return 1 + count(root.left) + count(root.right);
    }
}

C++ Code
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool btreeGameWinningMove(TreeNode* root, int n, int x) {
        auto node = dfs(root, x);
        int l = count(node->left), r = count(node->right);
        return max({l, r, n - l - r - 1}) > n / 2;
    }

    TreeNode* dfs(TreeNode* root, int x) {
        if (!root || root->val == x) {
            return root;
        }
        auto node = dfs(root->left, x);
        return node ? node : dfs(root->right, x);
    }

    int count(TreeNode* root) {
        if (!root) {
            return 0;
        }
        return 1 + count(root->left) + count(root->right);
    }
};

Go Code
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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func btreeGameWinningMove(root *TreeNode, n int, x int) bool {
	var dfs func(*TreeNode) *TreeNode
	dfs = func(root *TreeNode) *TreeNode {
		if root == nil || root.Val == x {
			return root
		}
		node := dfs(root.Left)
		if node != nil {
			return node
		}
		return dfs(root.Right)
	}

	var count func(*TreeNode) int
	count = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		return 1 + count(root.Left) + count(root.Right)
	}

	node := dfs(root)
	l, r := count(node.Left), count(node.Right)
	return max(max(l, r), n-l-r-1) > n/2
}

TypeScript Code
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/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function btreeGameWinningMove(root: TreeNode | null, n: number, x: number): boolean {
    const dfs = (root: TreeNode | null): TreeNode | null => {
        if (!root || root.val === x) {
            return root;
        }
        return dfs(root.left) || dfs(root.right);
    };

    const count = (root: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        return 1 + count(root.left) + count(root.right);
    };

    const node = dfs(root);
    const l = count(node.left);
    const r = count(node.right);
    return Math.max(l, r, n - l - r - 1) > n / 2;
}

JavaScript Code
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/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} n
 * @param {number} x
 * @return {boolean}
 */
var btreeGameWinningMove = function (root, n, x) {
    const dfs = root => {
        if (!root || root.val === x) {
            return root;
        }
        return dfs(root.left) || dfs(root.right);
    };

    const count = root => {
        if (!root) {
            return 0;
        }
        return 1 + count(root.left) + count(root.right);
    };

    const node = dfs(root);
    const l = count(node.left);
    const r = count(node.right);
    return Math.max(l, r, n - l - r - 1) > n / 2;
};