230. Kth Smallest Element in a BST

题目：

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).

链接：  http://leetcode.com/problems/kth-smallest-element-in-a-bst/

题解：

用了比较笨的方法 - inorder traversal，做了一下剪枝。

Time Complexity - O(k)， Space Complexity - O(k)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    List<Integer> nodeVal = new ArrayList<Integer>();

    public int kthSmallest(TreeNode root, int k) {
        if(root == null)
            return 0;
        inorder(root, k);
        return nodeVal.get(k - 1);
    }

    private void inorder(TreeNode root, int k) {
        if(nodeVal.size() >= k)
            return;
        if(root == null)
            return;
        inorder(root.left, k);
        nodeVal.add(root.val);
        inorder(root.right, k);

    }
}

Update:

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {

    private int count = 0;
    private int res = 0;
    public int kthSmallest(TreeNode root, int k) {
        if(root == null)
            return 0;
        inorder(root, k);
        return res;
    }

    private void inorder(TreeNode root, int k) {
        if(root == null)
            return;
        if(count >= k)
            return;
        inorder(root.left, k);
        if(count >= k)
            return;
        res = root.val;
        count++;
        inorder(root.right, k);

    }
}

Reference:

https://leetcode.com/discuss/43267/4-lines-in-c

https://leetcode.com/discuss/43464/what-if-you-could-modify-the-bst-nodes-structure

https://leetcode.com/discuss/43299/o-k-space-o-n-time-10-short-lines-3-solutions

https://leetcode.com/discuss/45684/share-my-c-iterative-alg

https://leetcode.com/discuss/43771/implemented-java-binary-search-order-iterative-recursive