230. Kth Smallest Element in a BST

Problem

Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes 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 <= 10^4

• 0 <= Node.val <= 10^4

  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?

Solution (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int k;
    private int count = 0;
    private int val;

    public int kthSmallest(TreeNode root, int k) {
        this.k = k;
        calculate(root);
        return val;
    }

    private void calculate(TreeNode node) {
        if (node.left == null && node.right == null) {
            count++;
            if (count == k) {
                this.val = node.val;
            }
            return;
        }
        if (node.left != null) {
            calculate(node.left);
        }
        count++;
        if (count == k) {
            this.val = node.val;
            return;
        }
        if (node.right != null) {
            calculate(node.right);
        }
    }
}

Solution (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} k
 * @return {number}
 */
var kthSmallest = function(root, k) {
    const nums = [];
    inorder(root, nums);
    return nums[k - 1];
};

function inorder(root, nums) {
    if (root === null) return;
    inorder(root.left, nums);
    nums.push(root.val);
    inorder(root.right, nums);
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).