[leetcode] 1261. Find Elements in a Contaminated Binary Tree

解柏
2023-12-01

Description

Given a binary tree with the following rules:

  1. root.val == 0
  2. If treeNode.val == x and treeNode.left != null, then treeNode.left.val == 2 * x + 1
  3. If treeNode.val == x and treeNode.right != null, then treeNode.right.val == 2 * x + 2
    Now the binary tree is contaminated, which means all treeNode.val have been changed to -1.

You need to first recover the binary tree and then implement the FindElements class:

  • FindElements(TreeNode* root) Initializes the object with a contamined binary tree, you need to recover it first.
  • bool find(int target) Return if the target value exists in the recovered binary tree.

Example 1:

Input
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
Output
[null,false,true]
Explanation
FindElements findElements = new FindElements([-1,null,-1]); 
findElements.find(1); // return False 
findElements.find(2); // return True 

Example 2:

Input
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
Output
[null,true,true,false]
Explanation
FindElements findElements = new FindElements([-1,-1,-1,-1,-1]);
findElements.find(1); // return True
findElements.find(3); // return True
findElements.find(5); // return False

Example 3:

Input
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
Output
[null,true,false,false,true]
Explanation
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

Constraints:

  • TreeNode.val == -1
  • The height of the binary tree is less than or equal to 20
  • The total number of nodes is between [1, 10^4]
  • Total calls of find() is between [1, 10^4]
  • 0 <= target <= 10^6

分析

题目的意思是:恢复Contaminated二叉树,实现find操作。思路也很直接,list递归存储恢复的二叉树的值。然后find的时候判断target是否在list集合里面就行了。我看了一下别人的实现,跟我的差不多,当然也有其他的神奇实现,读者自行研究哈哈哈。

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 solve(self,root,x):
        if(root is None):
            return 
        
        self.res.append(x)
        self.solve(root.left,2*x+1)
        self.solve(root.right,2*x+2)
        

    def __init__(self, root: TreeNode):
        self.res=[]
        self.solve(root,0)
        
    def find(self, target: int) -> bool:
        if(target in self.res):
            return True
        return False


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