[LeetCode] 230. Kth Smallest Element in a BST

Given the root of a binary search tree, and an integer k, return the kth (1-indexed) smallest element in the tree.

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

Constraints:

The number of nodes in the tree is n.
1 <= k <= n <= 104
0 <= Node.val <= 104

Follow up: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?

二叉搜索树中第 K 小的元素。

因为是 BST 所以大概率会考察中序遍历，即中序遍历的结果是升序的。这道题的最优解就是按照中序遍历的方法去遍历 BST 的节点，用 count 记录是否到 K，输出第 K 个节点即可。影子题671。

时间O(n)

空间O(n)

Java递归实现

 1 class Solution {
 2     private static int count;
 3     private static int res;
 4 
 5     public int kthSmallest(TreeNode root, int k) {
 6         count = k;
 7         helper(root);
 8         return res;
 9     }
10 
11     public void helper(TreeNode root) {
12         if (root == null) {
13             return;
14         }
15         helper(root.left);
16         count--;
17         if (count == 0) {
18             res = root.val;
19         }
20         helper(root.right);
21     }
22 }

Java迭代实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public int kthSmallest(TreeNode root, int k) {
18         Stack<TreeNode> stack = new Stack<>();
19         TreeNode cur = root;
20         stack.push(cur);
21         int count = 0;
22         while (cur != null || !stack.isEmpty()) {
23             while (cur != null) {
24                 stack.push(cur);
25                 cur = cur.left;
26             }
27             cur = stack.pop();
28             count++;
29             if (count == k) {
30                 return cur.val;
31             }
32             cur = cur.right;
33         }
34         return -1;
35     }
36 }

JavaScript递归实现

 1 /**
 2  * @param {TreeNode} root
 3  * @param {number} k
 4  * @return {number}
 5  */
 6 var kthSmallest = function (root, k) {
 7     let count = k;
 8     let res = 0;
 9 
10     let helper = function (root) {
11         if (root == null) {
12             return;
13         }
14         helper(root.left);
15         count--;
16         if (count == 0) {
17             res = root.val;
18         }
19         helper(root.right);
20     }
21     helper(root);
22     return res;
23 };

JavaScript迭代实现

 1 /**
 2  * @param {TreeNode} root
 3  * @param {number} k
 4  * @return {number}
 5  */
 6 var kthSmallest = function (root, k) {
 7     let stack = [];
 8     while (root != null || stack.length > 0) {
 9         if (root != null) {
10             stack.push(root);
11             root = root.left;
12         } else {
13             root = stack.pop();
14             k--;
15             if (k == 0) {
16                 return root.val;
17             }
18             root = root.right;
19         }
20     }
21     return -1;
22 };

CNoodle

Let the code speak for you.

[LeetCode] 230. Kth Smallest Element in a BST

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