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leetcode 1261. Find Elements in a Contaminated Binary Tree(python)

2022-01-29 15:45:08 Wang Daya

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describe

Given a binary tree with the following rules:

  • root.val == 0
  • If treeNode.val == x and treeNode.left != null, then treeNode.left.val == 2 * x + 1
  • 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.

Implement the FindElements class:

  • FindElements(TreeNode* root) Initializes the object with a contaminated binary tree and recovers it.
  • bool find(int target) Returns true 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 
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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
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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
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Note:

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
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analysis

According to the meaning , Given a tree , But you can only see the structure of the tree , Because the tree is polluted, the values of all nodes are -1 , However, the value of the node can be reproduced and restored , That is, the root node is 0 , The value of the left node is twice the value of its parent node plus one , The value of the right node is twice the value of its parent node . The title requires us to use __init__ Function first restores the tree , And then use find Function judgement target Whether it exists in the tree .

The idea is simple , Is to use recursion to directly calculate the values of the nodes of the tree and store them in a list , And then determine target Whether it is in the list . In fact, this problem looks complex , It's actually quite simple .

answer

class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
class FindElements(object):
    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.vals = []
        def dfs(root, val):
            if not root:
                return
            self.vals.append(val)
            if root.left:
                dfs(root.left, val*2+1)
            if root.right:
                dfs(root.right, val*2+2)
        dfs(root, 0)


    def find(self, target):
        """
        :type target: int
        :rtype: bool
        """
        if target in self.vals:
            return True
        return False

        	      
		
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Running results

Runtime: 622 ms, faster than 6.67% of Python online submissions for Find Elements in a Contaminated Binary Tree.
Memory Usage: 19.2 MB, less than 76.67% of Python online submissions for Find Elements in a Contaminated Binary Tree.
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analysis

You can also use queues to solve this problem , The process of building the tree is similar to the above process , I won't repeat .

answer

class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
class FindElements(object):
    def __init__(self, root):
        self.A = set()
        queue = collections.deque([[root,0]])
        while queue:
            n,x = queue.popleft()
            self.A.add(x)
            if n.left:
                queue.append( [n.left  , 2*x+1] )
            if n.right:
                queue.append( [n.right , 2*x+2] )
                
    def find(self, target):
        return target in self.A
        
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Running results

Runtime: 143 ms, faster than 33.33% of Python online submissions for Find Elements in a Contaminated Binary Tree.
Memory Usage: 19.7 MB, less than 13.33% of Python online submissions for Find Elements in a Contaminated Binary Tree.    
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Original link :leetcode.com/problems/fi…

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author[Wang Daya],Please bring the original link to reprint, thank you.
https://en.pythonmana.com/2022/01/202201291545062910.html

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