LeetCode 230. Kth Smallest Element in a BST

原题链接在这里：https://leetcode.com/problems/kth-smallest-element-in-a-bst/description/

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).

题解：

Method 1: BST 用Binary Tree Inorder Traversal出来的就是由小到大，Method1用recursion，会出来一整个list, 然后返回list.get(k-1)即可.

Time Complexity: O(n). Space: O(n).

AC Java:

/**
 * 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) {
        List<Integer> ls = new ArrayList<Integer>();
        inorder(root,ls);
        return ls.get(k-1);
    }
    private void inorder(TreeNode root, List<Integer> ls){
        if(root == null){
            return;
        }
        inorder(root.left,ls);
        ls.add(root.val);
        inorder(root.right,ls);
    }
}

Method 2: 改recursion为iteration, 如此可以不用走完全树.

Time Complexity: O(n). Space: O(logn).

/**
 * 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) {
        Stack<TreeNode> stk = new Stack<TreeNode>();
        while(root!=null || !stk.empty()){
            if(root != null){
                stk.push(root);
                root = root.left;
            }else{
                TreeNode tn = stk.pop();
                k--;
                if(k == 0){
                    return tn.val;
                }
                root = tn.right;
            }
        }
        return -1;
    }
}

Method 3: 根据BST性质，看root左子树有多少点，如果正好等于k-1, 则返回root.val, 如果小于k-1, 就在root右子树中找第k-leftSum-1小的点值, 若果大于k-1, 就在左子树中找第k小的点值。

Note: 是返回root.val 不是root.

Time Complexity: O(n). Space: O(logn).

AC Java:

/**
 * 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 leftSum = nodeSum(root.left);
        if(leftSum == k-1){
            return root.val;
        }else if(leftSum < k-1){
            return kthSmallest(root.right,k-leftSum-1);
        }else{
            return kthSmallest(root.left,k);
        }
    }
    private int nodeSum(TreeNode root){
        if(root == null){
            return 0;
        }
        return nodeSum(root.left)+nodeSum(root.right)+1;
    }
}