Skip to content

Latest commit

 

History

History
101 lines (86 loc) · 3.16 KB

File metadata and controls

101 lines (86 loc) · 3.16 KB

1261. 在受污染的二叉树中查找元素

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:

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

题解 (Python)

1. 题解

# 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)