[LeetCode] 1261. Find Elements in a Contaminated Binary Tree 在受汙染的二叉樹中查詢元素
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 alltreeNode.val
have been changed to-1
Implement theFindElements
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
這道題給了一棵二叉樹,由於被汙染了,所以每個結點值都是 -1,但是實際的命名規則是根結點值為0,且對於任意一個結點值x,若其左結點存在,則其值為 2x+1,若其右結點存在,則值為 2x+2,現在讓復原給定的二叉樹,同時對於給定的 target 值,判斷其是否在復原的二叉樹中。看了下題目中的條件,find 函式可能被呼叫上萬次,肯定不能每次呼叫都遍歷一遍二叉樹,最快速的查詢時間是常數級的,所以應該將所有的結點值都放到一個 HashSet 中,這樣就能最快速的查詢目標值了。這裡首先要做的就是復原二叉樹,在復原的過程中將結點值都存到 HashSet 中,可以用一個先序遍歷,傳入根結點值0。在遞迴函式中,若當前結點為空,直接返回,否則將傳入的 val 加入 HashSet,並且賦值給當前結點值。然後判斷,若左子結點存在,則對左子結點呼叫遞迴函式,並且將 2*val + 1
當作引數傳入,同理,若右子結點存在,則對右子結點呼叫遞迴函式,並且將 2*val + 2
當作引數傳入即可,參見程式碼如下:
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
參考資料:
https://leetcode.com/problems/find-elements-in-a-contaminated-binary-tree/
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