[LeetCode] 230. Kth Smallest Element in a BST

Problem

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.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
Output: 3

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?

Solution

Binary Search

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int countLeft = count(root.left);
        if (k <= countLeft) return kthSmallest(root.left, k);
        else if (k > countLeft+1) return kthSmallest(root.right, k-countLeft-1);
        else return root.val; //k == countLeft+1
    }
    private int count(TreeNode root) {
        int count = 0;
        if (root == null) return count;
        return count(root.left)+1+count(root.right);
    }
}

In-Order Traversal Iteration

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Deque stack = new ArrayDeque<>();
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
        while (k != 0) {
            TreeNode node = stack.pop();
            k--;
            if (k == 0) return node.val;
            TreeNode cur = node.right;
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
        }
        return -1;
    }
}

In-Order Traversal Recursion

class Solution {
    int res = 0;
    int count = 0;
    public int kthSmallest(TreeNode root, int k) {
        count = k;
        helper(root);
        return res;
    }
    private void helper(TreeNode root) {
        if (root == null) return;
        
        helper(root.left);
        
        count--;
        if (count == 0) res = root.val;
        
        helper(root.right);
    }
}