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[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所以大概率会碰到中序遍历inorder traversal，这一题也不例外。思路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(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public int kthSmallest(TreeNode root, int k) {
12         Stack<TreeNode> stack = new Stack<>();
13         while (root != null || !stack.isEmpty()) {
14             if (root != null) {
15                 stack.push(root);
16                 root = root.left;
17             } else {
18                 root = stack.pop();
19                 k--;
20                 if (k == 0) {
21                     return root.val;
22                 }
23                 root = root.right;
24             }
25         }
26         return Integer.MIN_VALUE;
27     }
28 }

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 };