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。
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
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
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
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
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class FindElements:
def __init__(self, root: TreeNode):
self.vals = set()
root.val = 0
stack = [root]
while stack:
curr = stack.pop()
self.vals.add(curr.val)
if curr.left:
curr.left.val = 2 * curr.val + 1
stack.append(curr.left)
if curr.right:
curr.right.val = 2 * curr.val + 2
stack.append(curr.right)
def find(self, target: int) -> bool:
return target in self.vals
# Your FindElements object will be instantiated and called as such:
# obj = FindElements(root)
# param_1 = obj.find(target)