README_EN.md

230. Kth Smallest Element in a BST

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Description

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?

Solutions

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def kthSmallest(self, root: TreeNode, k: int) -> int:
        def inorder(root):
            if root is None:
                return
            inorder(root.left)
            self.k -= 1
            if self.k == 0:
                self.res = root.val
                return
            inorder(root.right)
        self.k = k
        inorder(root)
        return self.res

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 res;

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

    private void inorder(TreeNode root) {
        if (root == null) {
            return;
        }
        inorder(root.left);
        if (--k == 0) {
            res = root.val;
            return;
        }
        inorder(root.right);
    }
}

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