1 """ 2 Given a binary tree with the following rules: 3 root.val == 0 4 If treeNode.val == x and treeNode.left != null, then treeNode.left.val == 2 * x + 1 5 If treeNode.val == x and treeNode.right != null, then treeNode.right.val == 2 * x + 2 6 Now the binary tree is contaminated, which means all treeNode.val have been changed to -1. 7 You need to first recover the binary tree and then implement the FindElements class: 8 FindElements(TreeNode* root) Initializes the object with a contamined binary tree, you need to recover it first. 9 bool find(int target) Return if the target value exists in the recovered binary tree. 10 Example 1: 11 Input 12 ["FindElements","find","find"] 13 [[[-1,null,-1]],[1],[2]] 14 Output 15 [null,false,true] 16 Explanation 17 FindElements findElements = new FindElements([-1,null,-1]); 18 findElements.find(1); // return False 19 findElements.find(2); // return True 20 Example 2: 21 Input 22 ["FindElements","find","find","find"] 23 [[[-1,-1,-1,-1,-1]],[1],[3],[5]] 24 Output 25 [null,true,true,false] 26 Explanation 27 FindElements findElements = new FindElements([-1,-1,-1,-1,-1]); 28 findElements.find(1); // return True 29 findElements.find(3); // return True 30 findElements.find(5); // return False 31 Example 3: 32 Input 33 ["FindElements","find","find","find","find"] 34 [[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]] 35 Output 36 [null,true,false,false,true] 37 Explanation 38 FindElements findElements = new FindElements([-1,null,-1,-1,null,-1]); 39 findElements.find(2); // return True 40 findElements.find(3); // return False 41 findElements.find(4); // return False 42 findElements.find(5); // return True 43 """ 44 class TreeNode: 45 def __init__(self, x): 46 self.val = x 47 self.left = None 48 self.right = None 49 50 class FindElements: 51 def __init__(self, root): 52 self.array = [] # !!!保存结点出现的值,注意self. 私有成员 53 if root != None: # 调用递归函数 54 self.recover_tree(root, 0) 55 def recover_tree(self, root, v): # 递归调用 56 root.val = v 57 self.array.append(v) 58 if root.left != None: 59 self.recover_tree(root.left, 2 * v + 1) 60 if root.right != None: 61 self.recover_tree(root.right, 2 * v + 2) 62 def find(self, target: int) -> bool: 63 return target in self.array # 再结果数组里找