Kth Smallest Element in a BST

https://leetcode.com/problems/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:

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

解题思路：

带返回值的递归

/**

 * 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) {

        int [] step = new int[1];

        step[0] = 1;

        TreeNode res = dfs(root, k, step);

        if(res == null) {

            return -1;

        }

        return res.val;

    }

    

    public TreeNode dfs(TreeNode root, int k, int[] step) {

        if(root == null) {

            return null;

        }

        TreeNode res1 = dfs(root.left, k, step);

        if(res1 != null) {

            return res1;

        }

        if(step[0] == k) {

            return root;

        }

        step[0]++;

        TreeNode res2 = dfs(root.right, k, step);

        return res2;

    }

}

不带返回值的递归

/**

 * 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) {

        int [] step = new int[1];

        step[0] = 1;

        int [] res = new int[1];

        dfs(root, k, step, res);

        return res[0];

    }

    

    public void dfs(TreeNode root, int k, int[] step, int[] res) {

        if(root == null) {

            return;

        }

        dfs(root.left, k, step, res);

        if(step[0] == k) {

            res[0] = root.val;

            step[0]++;

            return;

        }

        step[0]++;

        dfs(root.right, k, step, res);

    }

}

或者直接将step赋值为k也可以，省去一个参数

/**

 * 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) {

        int [] step = new int[1];

        step[0] = k;

        int [] res = new int[1];

        dfs(root, step, res);

        return res[0];

    }

    

    public void dfs(TreeNode root, int[] step, int[] res) {

        if(root == null) {

            return;

        }

        dfs(root.left, step, res);

        if(step[0] == 1) {

            res[0] = root.val;

            step[0]--;

            return;

        }

        step[0]--;

        dfs(root.right, step, res);

    }

}

参数使用数组的原因是Java无法pass by reference，或者使用全局变量，也是可以的。