LeetCode OJ 230. Kth Smallest Element in a BST
Total Accepted: 46445 Total Submissions: 122594 Difficulty: Medium
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.
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?
Hint:
- Try to utilize the property of a BST.
- What if you could modify the BST node's structure?
- The optimal runtime complexity is O(height of BST).
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
二叉搜索树特点是对于任意一个节点,它的左子树的节点值都比它小,它的右子树的值都比它大。因此可以从树的最左边向上递归,同时计数,直到计数达到当前指定数字。
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 private int cnt = 0; 12 public int kthSmallest(TreeNode root, int k) { 13 int result = 0; 14 if(root != null){ 15 result = kthSmallest(root.left,k); 16 cnt++; 17 if(cnt == k){ 18 return root.val; 19 } 20 result += kthSmallest(root.right,k); 21 } 22 return result; 23 } 24 }