标签：integer search pop class list roo note nbsp element
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.
Example 1:
Input: root = [3,1,4,null,2], k = 1 3 / 1 4 2 Output: 1
Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3
5
/ 3 6
/ 2 4
/
1
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?
class Solution { public int kthSmallest(TreeNode root, int k) { List<Integer> list = new ArrayList(); help(list, root); Collections.sort(list); return list.get(k - 1); } public void help(List<Integer> list, TreeNode root){ if(root == null) return; list.add(root.val); help(list, root.left); help(list, root.right); } }
遍历，存到list，sort，然后get
public class Solution { public int kthSmallest(TreeNode root, int k) { Stack<TreeNode> s = new Stack<>(); TreeNode p = root; while (!s.empty() || p != null) { if (p != null) { s.push(p); p = p.left; } else { p = s.pop(); --k; if (k == 0) { return p.val; } p = p.right; } } return -1; } }
中序（左根右遍历）存入stack，
230. Kth Smallest Element in a BST
标签：integer search pop class list roo note nbsp element
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11451776.html
