230. Kth Smallest Element in a BST

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note: 
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

Hint:

1. Try to utilize the property of a BST.
2. What if you could modify the BST node's structure?
3. The optimal runtime complexity is O(height of BST).

Similar: Binary Tree Inorder Traversal

Solution 1. Binary Search
Time Complexity: T(n) = n/2 + T(n/2) = O(n)
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int count = countNodes(root.left);
        if (k <= count) {
            return kthSmallest(root.left, k);
        } else if (k > count + 1) {
            return kthSmallest(root.right, k-1-count); 
        }
        return root.val;
    }

    public int countNodes(TreeNode n) {
        if (n == null) return 0;
        return 1 + countNodes(n.left) + countNodes(n.right);
    }
}

Solution 2. Modify the structure of BST
Time Complexity: O(lg n)
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        TreeWithCount twcRoot = buildTWC(root);
        return findKth(twcRoot, k);
    }
    
    private int findKth(TreeWithCount root, int k) {
        if (k <= 0 || k > root.count) return -1;
        if (root.left != null) {
            if (root.left.count == k-1) return root.val;
            if (k <= root.left.count) return findKth(root.left, k);
            return findKth(root.right, k-1-root.left.count);
        } else {
            if (k == 1) return root.val;
            return findKth(root.right, k-1);
        }
    }
    
    private TreeWithCount buildTWC(TreeNode root) {
        if (root == null) return null;
        
        TreeWithCount twcRoot = new TreeWithCount(root.val);
        twcRoot.left = buildTWC(root.left);
        twcRoot.right = buildTWC(root.right);
        if (twcRoot.left != null) twcRoot.count += twcRoot.left.count;
        if (twcRoot.right != null) twcRoot.count += twcRoot.right.count;
        return twcRoot;
    }
    
    class TreeWithCount {
        int val;
        int count;
        TreeWithCount left;
        TreeWithCount right;
        public TreeWithCount (int val) {
            this.val = val;
            this.count = 1;
        }
    } 
}