[LeetCode] 1261. Find Elements in a Contaminated Binary Tree 在受污染的二叉树中查找元素
Given a binary tree with the following rules:
root.val == 0
- If
treeNode.val == x
andtreeNode.left != null
, thentreeNode.left.val == 2 * x + 1
- If
treeNode.val == x
andtreeNode.right != null
, thentreeNode.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)
Returnstrue
if thetarget
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 <= 106
这道题给了一棵二叉树,由于被污染了,所以每个结点值都是 -1,但是实际的命名规则是根结点值为0,且对于任意一个结点值x,若其左结点存在,则其值为 2x+1,若其右结点存在,则值为 2x+2,现在让复原给定的二叉树,同时对于给定的 target 值,判断其是否在复原的二叉树中。看了下题目中的条件,find 函数可能被调用上万次,肯定不能每次调用都遍历一遍二叉树,最快速的查找时间是常数级的,所以应该将所有的结点值都放到一个 HashSet 中,这样就能最快速的查找目标值了。这里首先要做的就是复原二叉树,在复原的过程中将结点值都存到 HashSet 中,可以用一个先序遍历,传入根结点值0。在递归函数中,若当前结点为空,直接返回,否则将传入的 val 加入 HashSet,并且赋值给当前结点值。然后判断,若左子结点存在,则对左子结点调用递归函数,并且将 2*val + 1
当作参数传入,同理,若右子结点存在,则对右子结点调用递归函数,并且将 2*val + 2
当作参数传入即可,参见代码如下:
class FindElements {
public:
FindElements(TreeNode* root) {
helper(root, 0);
}
bool find(int target) {
return st.count(target);
}
private:
unordered_set<int> st;
void helper(TreeNode* node, int val) {
if (!node) return;
st.insert(val);
node->val = val;
if (node->left) helper(node->left, 2 * val + 1);
if (node->right) helper(node->right, 2 * val + 2);
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1261
参考资料:
https://leetcode.com/problems/find-elements-in-a-contaminated-binary-tree/