平衡二叉树中第k小的数 Kth Smallest Element in a BST

问题：

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?

解决：

① 借助中序遍历。

class Solution {//3ms
    public int kthSmallest(TreeNode root, int k) {
        List<Integer> list = new ArrayList<>();
        inorder(root,list);
        int res = list.get(k - 1);
        return res;
    }
    public void inorder(TreeNode root,List<Integer> list){
        if (root == null) return;
        inorder(root.left,list);
        list.add(root.val);
        inorder(root.right,list);
    }
}

② 借助栈实现中序遍历。

class Solution {//2ms
    public int kthSmallest(TreeNode root, int k) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        while (cur != null){
            stack.push(cur);
            cur = cur.left;
        }
        int count = 0;
        while (! stack.isEmpty()){
            cur = stack.pop();
            count ++;
            if (count == k){
                return cur.val;
            }
            TreeNode tmp = cur.right;
            while (tmp != null){
                stack.push(tmp);
                tmp = tmp.left;
            }
        }
        return -1;
    }
}

③ 在中序遍历过程中查找第k小的值。

class Solution { //0ms
    int count = 0;
    int res = 0;
    public int kthSmallest(TreeNode root, int k) {
        count = k - 1;
        inorder(root);
        return res;
    }
    public void inorder(TreeNode root){
        if (root == null || count < 0){
            return;
        }
        inorder(root.left);
        if (count == 0){
            res = root.val;
            count --;
        }
        count --;
        inorder(root.right);
        
    }
}