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：

重点是，利用BST的中序遍历是一个有序序列，再从中寻找第K小的值。

注意的是，step什么时候+1？显然是在遍历到每个节点的时候。

那么在那么多递归中，实际上只有下面一段才是真正的执行，dfs left和right都不需要+1。

if (step[0] == k) {
return root;
}
step[0]++;

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。