Forgetting null/base-case handling
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.
Move from brute-force thinking to an efficient approach using tree strategy.
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:
n.1 <= x <= n <= 100n is odd.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[1,2,3,4,5,6,7,8,9,10,11] 11 3
[1,2,3] 3 1
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1145: Binary Tree Coloring Game
/**
* 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);
}
}
// Accepted solution for LeetCode #1145: Binary Tree Coloring Game
/**
* 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
}
# Accepted solution for LeetCode #1145: Binary Tree Coloring Game
# 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
// Accepted solution for LeetCode #1145: Binary Tree Coloring Game
struct Solution;
use rustgym_util::*;
trait Postorder {
fn postorder(&self, x: i32, left: &mut usize, right: &mut usize) -> usize;
}
impl Postorder for TreeLink {
fn postorder(&self, x: i32, left: &mut usize, right: &mut usize) -> usize {
if let Some(node) = self {
let node = node.borrow();
let val = node.val;
let l = node.left.postorder(x, left, right);
let r = node.right.postorder(x, left, right);
if val == x {
*left = l;
*right = r;
}
l + r + 1
} else {
0
}
}
}
impl Solution {
fn btree_game_winning_move(root: TreeLink, n: i32, x: i32) -> bool {
let mut left = 0;
let mut right = 0;
root.postorder(x, &mut left, &mut right);
let n = n as usize;
let top = n - 1 - left - right;
top.max(left.max(right)) > n / 2
}
}
#[test]
fn test() {
let root = tree!(
1,
tree!(
2,
tree!(4, tree!(8), tree!(9)),
tree!(5, tree!(10), tree!(11))
),
tree!(3, tree!(6), tree!(7))
);
let n = 11;
let x = 3;
let res = true;
assert_eq!(Solution::btree_game_winning_move(root, n, x), res);
}
// Accepted solution for LeetCode #1145: Binary Tree Coloring Game
/**
* 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;
}
Use this to step through a reusable interview workflow for this problem.
BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.
Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.
Review these before coding to avoid predictable interview regressions.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.