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 <= 100nis odd.- 1 <= Node.val <= n
- All the values of the tree are unique.
# 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)
t = n - l - r - 1
m = max(l, r, t)
return m > n - m/**
* 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);
int m = Math.max(Math.max(l, r), n - l - r - 1);
return m > n - m;
}
private int count(TreeNode node) {
if (node == null) {
return 0;
}
return 1 + count(node.left) + count(node.right);
}
private TreeNode dfs(TreeNode root, int x) {
if (root == null || root.val == x) {
return root;
}
TreeNode l = dfs(root.left, x);
if (l != null) {
return l;
}
return dfs(root.right, x);
}
}/**
* 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) {
TreeNode* node = dfs(root, x);
int l = count(node->left);
int r = count(node->right);
int m = max(max(l, r), n - l - r - 1);
return m > n - m;
}
int count(TreeNode* root) {
if (!root) return 0;
return 1 + count(root->left) + count(root->right);
}
TreeNode* dfs(TreeNode* root, int x) {
if (!root || root->val == x) return root;
auto l = dfs(root->left, x);
return l ? l : dfs(root->right, x);
}
};/**
* 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
}
l := dfs(root.Left)
if l != nil {
return l
}
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)
m := max(max(l, r), n-l-r-1)
return m > n-m
}
func max(a, b int) int {
if a > b {
return a
}
return b
}