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

### 描述

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

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

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

### 解答

``````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

### 运行结果

``````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.

### 解答

``````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()
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

### 运行结果

``````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.