给出一个满足下述规则的二叉树:
root.val == 0
- 如果
treeNode.val == x
且treeNode.left != null
,那么treeNode.left.val == 2 * x + 1
- 如果
treeNode.val == x
且treeNode.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:
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:
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:
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
类
Java
java
/**
* 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
# 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
/**
* 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)
*/