Leetcode: 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:

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

Java Solution 1 - Inorder Traversal

We can inorder traverse the tree and get the kth smallest element. Time is 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) {
        TreeNode node = root;
        Stack<TreeNode> st = new Stack<TreeNode>();
        int counter = 0;
        while (!st.isEmpty() || node != null) {
            if (node != null) {
                st.push(node);
                node = node.left;
            } 
            else {
                node = st.pop();
                counter++;
                if (counter == k) return node.val;
                node = node.right;
            }
        }
        return -1;
    }
}

Recursion method:

public class Solution {
    int count = 0;
    
    public int kthSmallest(TreeNode root, int k) {
        List<Integer> res = new ArrayList<Integer>();
        res.add(null);
        helper(root, k, res);
        return res.get(0);
    }
    
    public void helper(TreeNode root, int k, List<Integer> res) {
        if (root == null) return;
        helper(root.left, k, res);
        count++;
        if (count == k) res.set(0, root.val);
        helper(root.right, k, res);
    }
}

Java Solution 2  Binary Search

We can let each node track the order, i.e., the number of elements that are less than itself(left Subtree size). Time is O(log(n)).

当前做法是O(NlogN)

如果我们频繁的操作该树，并且频繁的调用kth函数，有什么优化方法使时间复杂度降低至O(h)？h是树的高度。根据提示，我们可以在TreeNode中加入一个rank成员，这个变量记录的是该节点的左子树中节点的个数，其实就是有多少个节点比该节点小。这样我们就可以用二叉树搜索的方法来解决这个问题了。这个添加rank的操作可以在建树的时候一起完成。

  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); // 1 is counted as current node
        }

        return root.val;
    }

    public int countNodes(TreeNode n) {
        if (n == null) return 0;

        return 1 + countNodes(n.left) + countNodes(n.right);
    }