Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.
Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
Example 1:
Input: root = [3,1,4,null,2], k = 1 Output: 1
Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3 Output: 3
Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
<Solution>What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
想法如下
- 因為 BST 的特性,用 inorder 歷遍的話,數字剛好會是從最小到最大
- 檢查目前的節點,是不是剛好就是指定的第 k 個
Java(參考解答)
kotlin
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