Last updated on October 25th, 2024 at 10:21 pm

Here, We see Kth Smallest Element in a BST LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Kth Smallest Element in a BST LeetCode Solution

Problem Statement

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

1. Kth Smallest Element in a BST LeetCode Solution C++

class Solution {
public:
    void preOrderTraversal(TreeNode* root, vector<int> &v){
        if(root == NULL)    return;
        v.push_back(root->val);
        preOrderTraversal(root->left, v);
        preOrderTraversal(root->right, v);      
    }
    int kthSmallest(TreeNode* root, int k) {
        vector<int> v; 
        preOrderTraversal(root, v);
        sort(v.begin(), v.end());
        return v[k-1];
    }
};

2. Kth Smallest Element in a BST LeetCode Solution Java

class Solution {
    List<Integer> values = new ArrayList<>();
    public int kthSmallest(TreeNode root, int k) {
        inorder(root);
        return values.get(k - 1);
    }

    private void inorder(TreeNode root) {
        if (root == null) {
            return;
        }
        inorder(root.left);
        values.add(root.val);
        inorder(root.right);
    }
}

3. Kth Smallest Element in a BST Solution JavaScript

var kthSmallest = function(root, k) {
    let vals = [];
    (function dfs(node) {
        if (vals.length !=k) {
        if(node.left) dfs(node.left);
        vals.push(node.val);
        if (node.right) dfs(node.right);
        }  
    })(root)
    return vals[k-1]    
};

4. Kth Smallest Element in a BST Solution Python

class Solution(object):
    def kthSmallest(self, root, k):
        values = []
        self.inorder(root, values)
        return values[k - 1]

    def inorder(self, root, values):
        if root is None:
            return
        self.inorder(root.left, values)
        values.append(root.val)
        self.inorder(root.right, values)