在受污染的二叉树中查找元素(Lc1261)——DFS+哈希表

给出一个满足下述规则的二叉树:

  1. root.val == 0
  2. 如果 treeNode.val == xtreeNode.left != null,那么 treeNode.left.val == 2 * x + 1
  3. 如果 treeNode.val == xtreeNode.right != null,那么 treeNode.right.val == 2 * x + 2

现在这个二叉树受到「污染」,所有的 treeNode.val 都变成了 -1

请你先还原二叉树,然后实现 FindElements 类:

  • FindElements(TreeNode* root) 用受污染的二叉树初始化对象,你需要先把它还原。
  • bool find(int target) 判断目标值 target 是否存在于还原后的二叉树中并返回结果。

示例 1:

复制代码
输入:
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
输出:
[null,false,true]
解释:
FindElements findElements = new FindElements([-1,null,-1]); 
findElements.find(1); // return False 
findElements.find(2); // return True 

示例 2:

复制代码
输入:
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
输出:
[null,true,true,false]
解释:
FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
findElements.find(1); // return True
findElements.find(3); // return True
findElements.find(5); // return False

示例 3:

复制代码
输入:
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
输出:
[null,true,false,false,true]
解释:
FindElements findElements = new FindElements([-1,null,-1,-1,null,-1]);
findElements.find(2); // return True
findElements.find(3); // return False
findElements.find(4); // return False
findElements.find(5); // return True

提示:

  • TreeNode.val == -1
  • 二叉树的高度不超过 20
  • 节点的总数在 [1, 10^4] 之间
  • 调用 find() 的总次数在 [1, 10^4] 之间
  • 0 <= target <= 10^6

问题简要描述:还原二叉树并实现 FindElements

Java

java 复制代码
/**
 * 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 FindElements {
    Set<Integer> set = new HashSet<>();
    public FindElements(TreeNode root) {
        root.val = 0;
        dfs(root);
    }

    public boolean find(int target) {
        return set.contains(target);
    }

    void dfs(TreeNode root) {
        set.add(root.val);
        if (root.left != null) {
            root.left.val = 2 * root.val + 1;
            dfs(root.left);
        }
        if (root.right != null) {
            root.right.val = 2 * root.val + 2;
            dfs(root.right);
        }
    }
}

/**
 * Your FindElements object will be instantiated and called as such:
 * FindElements obj = new FindElements(root);
 * boolean param_1 = obj.find(target);
 */

Python3

python 复制代码
# 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 FindElements:

    def __init__(self, root: Optional[TreeNode]):
        def dfs(root: Optional[TreeNode]):
            self.s.add(root.val)
            if root.left:
                root.left.val = root.val * 2 + 1
                dfs(root.left)
            if root.right:
                root.right.val = root.val * 2 + 2
                dfs(root.right)
        self.s = set()
        root.val = 0
        dfs(root)

    def find(self, target: int) -> bool:
        return target in self.s        



# Your FindElements object will be instantiated and called as such:
# obj = FindElements(root)
# param_1 = obj.find(target)

TypeScript

TypeScript 复制代码
/**
 * 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)
 *     }
 * }
 */

class FindElements {
    private s = new Set();
    constructor(root: TreeNode | null) {
        const dfs = (root: TreeNode | null) => {
            this.s.add(root.val)
            if (root.left != null) {
                root.left.val = root.val * 2 + 1;
                dfs(root.left);
            }
            if (root.right != null) {
                root.right.val = root.val * 2 + 2;
                dfs(root.right);
            }
        }
        root.val = 0;
        dfs(root);        
    }

    find(target: number): boolean {
        return this.s.has(target);    
    }
}

/**
 * Your FindElements object will be instantiated and called as such:
 * var obj = new FindElements(root)
 * var param_1 = obj.find(target)
 */
相关推荐
OrangeJiuce36 分钟前
【QT中的一些高级数据结构,持续更新中...】
数据结构·c++·qt
学编程的小程4 小时前
LeetCode216
算法·深度优先
leeyayai_xixihah4 小时前
2.21力扣-回溯组合
算法·leetcode·职场和发展
01_4 小时前
力扣hot100——相交,回文链表
算法·leetcode·链表·双指针
萌の鱼4 小时前
leetcode 2826. 将三个组排序
数据结构·c++·算法·leetcode
Buling_04 小时前
算法-哈希表篇08-四数之和
数据结构·算法·散列表
AllowM4 小时前
【LeetCode Hot100】除自身以外数组的乘积|左右乘积列表,Java实现!图解+代码,小白也能秒懂!
java·算法·leetcode
RAN_PAND4 小时前
STL介绍1:vector、pair、string、queue、map
开发语言·c++·算法
左灯右行的爱情6 小时前
Redis数据结构总结-listPack
数据结构·数据库·redis
fai厅的秃头姐!6 小时前
C语言03
c语言·数据结构·算法