1261. 在受汙染的二叉樹中查詢元素
給出一個滿足下述規則的二叉樹:
root.val == 0
如果 treeNode.val == x 且treeNode.left != null,那麼treeNode.left.val == 2 * x + 1
如果 treeNode.val == x 且 treeNode.right != null,那麼treeNode.right.val == 2 * x + 2
現在這個二叉樹受到「汙染」,所有的treeNode.val都變成了-1。
請你先還原二叉樹,然後實現FindElements類:
FindElements(TreeNode* root)用受汙染的二叉樹初始化物件,你需要先把它還原。
bool find(int target)判斷目標值target是否存在於還原後的二叉樹中並返回結果。
示例 1:
輸入:
["FindElements","find","find"]
[[[-1,null,-1]],[1],[2]]
輸出:
[null,false,true]
解釋:
FindElements findElements = new FindElements([-1,null,-1]);
findElements.find(1); // return False
findElements.find(2); // return True
示例 2:
輸入:
["FindElements","find","find","find"]
[[[-1,-1,-1,-1,-1]],[1],[3],[5]]
輸出:
解釋:
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:
輸入:
["FindElements","find","find","find","find"]
[[[-1,null,-1,-1,null,-1]],[2],[3],[4],[5]]
輸出:
[null,true,false,false,true]
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
二叉樹的高度不超過20
節點的總數在[1,10^4]之間
呼叫find()的總次數在[1,10^4]之間
0 <= target <= 10^6
來源:力扣(LeetCode)
連結:https://leetcode-cn.com/problems/find-elements-in-a-contaminated-binary-tree
很容易想到遞迴建樹,然後把值放在set()中,再遞迴查詢
# 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.vis=set() def recover(node): if not node: return node if node.left: node.left.val=2*node.val+1 self.vis.add(node.left.val) if node.right: node.right.val=2*node.val+2 self.vis.add(node.right.val) recover(node.left) recover(node.right) return node root.val=0 self.vis.add(0) self.node=recover(root) def find(self, target: int) -> bool: return target in self.vis # Your FindElements object will be instantiated and called as such: # obj = FindElements(root) # param_1 = obj.find(target)
然而
我就盲猜更好的解法是用二進位制
然後猜不下去了
如果我們把樹中的數全部加 1
發現每一行的左右子樹分別有不同的字首:
發現0是向左 , 1是向右
# 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): def recover(node): if not node: return node if node.left: node.left.val=2*node.val+1 if node.right: node.right.val=2*node.val+2 recover(node.left) recover(node.right) return node root.val=0 self.node=recover(root) def find(self, target: int) -> bool: node=self.node for binary in bin(target+1)[3:]: node=node and (node.left,node.right)[int(binary)] return bool(node) # Your FindElements object will be instantiated and called as such: # obj = FindElements(root) # param_1 = obj.find(target)
然後
。。。
分析一下複雜度,解法一時間複雜度 O(1),解法二O(1)-Space-O(logn)-Time
總結:解法一空間換時間,解法二可以預防MLE