**Kth Smallest in a Binary Search Tree - Amazon Top Interview Questions**

### Problem Statement :

Given a binary search tree root, and k return the kth (0-indexed) smallest value in root. It is guaranteed that the tree has at least k + 1 nodes. Constraints k ≤ n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [3, [2, null, null], [9, [7, [4, null, null], [8, null, null]], [12, null, null]]] k = 2 Output 4 Example 2 Input root = [3, [2, null, null], [9, [7, [4, null, null], [8, null, null]], [12, null, null]]] k = 0 Output 2

### Solution :

` ````
Solution in C++ :
int ans = 0;
void helper(Tree* root, int& k) {
if (root) {
helper(root->left, k);
if (k == 0) ans = root->val;
k--;
helper(root->right, k);
}
}
int solve(Tree* root, int k) {
helper(root, k);
return ans;
}
```

` ````
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
public int solve(Tree root, int k) {
ArrayList<Integer> al = new ArrayList<>();
helper(root, al);
return al.get(k);
}
public void helper(Tree node, ArrayList<Integer> al) {
if (node == null)
return;
helper(node.left, al);
al.add(node.val);
helper(node.right, al);
}
}
```

` ````
Solution in Python :
# class Tree:
# def __init__(self, val, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
def inorder(self):
if self.left:
for node in inorder(self.left):
yield node
yield self
if self.right:
for node in inorder(self.right):
yield node
setattr(Tree, "inorder", inorder)
class Solution:
def solve(self, root, k):
for node in root.inorder():
if k == 0:
return node.val
k -= 1
```

## View More Similar Problems

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View Solution →