Kth Smallest Element in a BST
Given the root of a binary search tree and an integer k, return the k-th smallest value (1-indexed) in the BST.
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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.
Input
The root of a BST and an integer `k`.
Output
The k-th smallest value in the BST.
Examples
Input: root = [3,1,4,null,2], k = 1
Output: 1
Inorder: [1,2,3,4]. First smallest is 1.
Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3
Inorder: [1,2,3,4,5,6]. Third smallest is 3.
The brute-force approach
Collect all values via inorder traversal into an array. Return array[k-1].
vals = []
def inorder(node):
if not node: return
inorder(node.left)
vals.append(node.val)
inorder(node.right)
inorder(root)
return vals[k-1]O(n) time and space. Stores all values even though we only need the k-th.
Spotting the pattern
This is a Tree Traversal problem. The key question to ask yourself:
Why does inorder traversal (left → node → right) of a BST produce values in sorted order? What property of a BST makes this true?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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