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In-order traversal of a BST yields sorted values, which is the key to solving this.
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.
Perform In-Order traversal (Left, Root, Right). Keep a counter. When counter reaches k, return the current node value.
Time complexity is O(H + k). Space complexity is O(H).
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