leetcode230- Kth Smallest Element in a BST- medium

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?

利用BST的中序遍历就是排序数组的性质。找第k大，那就中序遍历+cnt，当cnt到了第k个的时候停止遍历即可。中序遍历要点（node和stack条件或；左推到底；吐一个；等于右）

Follow up：在树的设计里维护一个int记录节点下面结点总个数。找第k大用递归。当左结点个数>=k大，那就在左子树找；当左节点个数== k-1<那就是当前结点的值；当左结点个数 < k -1，那就去右子树找。（另外最前面注意检查下左结点是不是null，是的话那就直接找当前或右子树。）

实现本题：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        int cnt = 0;
        TreeNode node = root;
        Stack<TreeNode> stack = new Stack<>();
        
        int result = 0;
        while (cnt < k) {
            while (node != null) {
                stack.push(node);
                node = node.left;
            }
            node = stack.pop();
            result = node.val;
            cnt++;
            node = node.right;
        }
        return result;
    }
}

实现follow up：(credit to leetcode discussion)

 public class Solution {
        public int kthSmallest(TreeNode root, int k) {
            TreeNodeWithCount rootWithCount = buildTreeWithCount(root);
            return kthSmallest(rootWithCount, k);
        }
        
        private TreeNodeWithCount buildTreeWithCount(TreeNode root) {
            if (root == null) return null;
            TreeNodeWithCount rootWithCount = new TreeNodeWithCount(root.val);
            rootWithCount.left = buildTreeWithCount(root.left);
            rootWithCount.right = buildTreeWithCount(root.right);
            if (rootWithCount.left != null) rootWithCount.count += rootWithCount.left.count;
            if (rootWithCount.right != null) rootWithCount.count += rootWithCount.right.count;
            return rootWithCount;
        }
        
        private int kthSmallest(TreeNodeWithCount rootWithCount, int k) {
            if (k <= 0 || k > rootWithCount.count) return -1;
            if (rootWithCount.left != null) {
                if (rootWithCount.left.count >= k) return kthSmallest(rootWithCount.left, k);
                if (rootWithCount.left.count == k-1) return rootWithCount.val;
                return kthSmallest(rootWithCount.right, k-1-rootWithCount.left.count);
            } else {
                if (k == 1) return rootWithCount.val;
                return kthSmallest(rootWithCount.right, k-1);
            }
        }
        
        class TreeNodeWithCount {
            int val;
            int count;
            TreeNodeWithCount left;
            TreeNodeWithCount right;
            TreeNodeWithCount(int x) {val = x; count = 1;};
        }
    }