Leetcode 230-Kth Smallest Element in a BST-Medium

题目：

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

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?

思路：

BST的重要性质：中序遍历（inorder）BST的结果：值从小到大输出所有node

本题k的最小值为1.

解法1: 迭代+栈 解法2: 递归

解法1: 非递归 （iterativaly）

中序遍历: 用栈存储：

循环1：从头结点开始，循环2（一路向左把所有左结点压入栈中，当把最左侧的结点压进去后），pop栈顶元素，并把当前pop结点的右子树压入栈。直到栈为空。

由于要找第Kth小的结点的值，也就是第K次pop出来的结点，所以定义一个全局变量count = 0, 每次pop一个出来，count++, 当count == k的时候，return root.val

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int count = 0;
    public int kthSmallest(TreeNode root, int k) {
        return inorder(root, k);
        
    }
    
    private int inorder(TreeNode root, int k) {
        Stack<TreeNode> s = new Stack<>();
        
        while(root != null || !s.isEmpty()) {
            while(root != null) {
                s.add(root);
                root = root.left;
            }
            root = s.pop();
            if(++count == k) return root.val;
            root = root.right;
        }
        // 没有找到（root == null || K < 1 || k > BST's total elements.）
        return -1;
    }
}

解法二：递归法

中序遍历采用递归法，代码简介很多。如果有左子树就继续递归，否则就处理当前结点，再递归右子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int count = 0;
    private int res = -1;
    
    public int kthSmallest(TreeNode root, int k) {
        inorder(root, k);
        return res;
    }
    
    private void inorder(TreeNode root, int k) {
        if(root.left != null) {
            inorder(root.left, k);
        }
        if(++count == k) {
            res = root.val;
            return;
        }
        if(root.right != null) {
            inorder(root.right, k);
        }
    }
}