【Leetcode】Kth Smallest Element in a BST

题目链接：https://leetcode.com/problems/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).

思路：

1、BST中序遍历的结果是对所有结点值逐渐增大的一个排列。用List保存遍历结果，返回List的第k个元素就行了。时间复杂度是O(n)，空间复杂度是O(n)。

2、中序遍历，用计数器记录当前遍历的结点个数，当遍历到k个元素的时候返回该结点的val，无需保存遍历的结点，时间复杂度O(n)，空间复杂度O(1)。

3、修改结点数据结构，在TreeNode结点加入leftCount，表示左孩子数目，根据结点的leftCount调整递归分支，它不会遍历所有结点，而是自上而下的路径，所以时间复杂度是O(height of BST)。

算法1：

	public int kthSmallest(TreeNode root, int k) {
		List<Integer> list = new ArrayList<Integer>();
		list = inOrder(root, list);
		return list.get(k - 1);
	}

	public List<Integer> inOrder(TreeNode p, List<Integer> order) {
		if (p != null) {
			order = inOrder(p.left, order);
			order.add(p.val);
			order = inOrder(p.right, order);
		}
		return order;
	}

算法2：

	int count = 0, result;

	public int kthSmallest(TreeNode root, int k) {
		if (root != null) {
			kthSmallest(root.left, k);
			count++;
			if (count == k) {
				result = root.val;
			}
			kthSmallest(root.right, k);
		}
		return result;
	}

算法3：

class newTreeNode { // BST
	newTreeNode left, right;
	int val, leftCount;// 记录左孩子个数

	newTreeNode(int val) {
		this.val = val;
	}
}

	int result;

	public int kthSmallest(newTreeNode root, int k) {
		if (root != null) {
			if (root.leftCount + 1 == k) {
				result = root.val;
			} else if (root.leftCount + 1 < k) {
				kthSmallest(root.right, k - (root.leftCount + 1));
			} else {
				kthSmallest(root.left, k);
			}
		}
		return result;
	}