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1145. 二叉树着色游戏

题目描述

有两位极客玩家参与了一场「二叉树着色」的游戏。游戏中,给出二叉树的根节点 root,树上总共有 n 个节点,且 n 为奇数,其中每个节点上的值从 1 到 n 各不相同。

最开始时:

  • 「一号」玩家从 [1, n] 中取一个值 x1 <= x <= n);
  • 「二号」玩家也从 [1, n] 中取一个值 y1 <= y <= n)且 y != x

「一号」玩家给值为 x 的节点染上红色,而「二号」玩家给值为 y 的节点染上蓝色。

之后两位玩家轮流进行操作,「一号」玩家先手。每一回合,玩家选择一个被他染过色的节点,将所选节点一个 未着色 的邻节点(即左右子节点、或父节点)进行染色(「一号」玩家染红色,「二号」玩家染蓝色)。

如果(且仅在此种情况下)当前玩家无法找到这样的节点来染色时,其回合就会被跳过。

若两个玩家都没有可以染色的节点时,游戏结束。着色节点最多的那位玩家获得胜利 ✌️。

现在,假设你是「二号」玩家,根据所给出的输入,假如存在一个 y 值可以确保你赢得这场游戏,则返回 true ;若无法获胜,就请返回 false

 

示例 1 :

输入:root = [1,2,3,4,5,6,7,8,9,10,11], n = 11, x = 3
输出:true
解释:第二个玩家可以选择值为 2 的节点。

示例 2 :

输入:root = [1,2,3], n = 3, x = 1
输出:false

 

提示:

  • 树中节点数目为 n
  • 1 <= x <= n <= 100
  • n 是奇数
  • 1 <= Node.val <= n
  • 树中所有值 互不相同

解法

方法一:DFS

我们先通过 $DFS$,找到「一号」玩家着色点 $x$ 所在的节点,记为 $node$。

接下来,我们统计 $node$ 的左子树、右子树的节点个数,分别记为 $l$ 和 $r$,而 $node$ 父节点方向上的个数为 $n - l - r - 1$。只要满足 $\max(l, r, n - l - r - 1) > \frac{n}{2}$,则「二号」玩家存在一个必胜策略。

时间复杂度 $O(n)$,空间复杂度 $O(n)$。其中 $n$ 是节点总数。

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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
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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);
    }
}
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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);
    }
};
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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
}
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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;
}
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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;
};

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