Leetcode: 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:

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

Java Solution 1 - Inorder Traversal

We can inorder traverse the tree and get the kth smallest element. Time is O(n).

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int kthSmallest(TreeNode root, int k) {
12         TreeNode node = root;
13         Stack<TreeNode> st = new Stack<TreeNode>();
14         int counter = 0;
15         while (!st.isEmpty() || node != null) {
16             if (node != null) {
17                 st.push(node);
18                 node = node.left;
19             } 
20             else {
21                 node = st.pop();
22                 counter++;
23                 if (counter == k) return node.val;
24                 node = node.right;
25             }
26         }
27         return -1;
28     }
29 }

Recursion method:

 1 public class Solution {
 2     int count = 0;
 3     
 4     public int kthSmallest(TreeNode root, int k) {
 5         List<Integer> res = new ArrayList<Integer>();
 6         res.add(null);
 7         helper(root, k, res);
 8         return res.get(0);
 9     }
10     
11     public void helper(TreeNode root, int k, List<Integer> res) {
12         if (root == null) return;
13         helper(root.left, k, res);
14         count++;
15         if (count == k) res.set(0, root.val);
16         helper(root.right, k, res);
17     }
18 }

Java Solution 2 Binary Search

We can let each node track the order, i.e., the number of elements that are less than itself(left Subtree size). Time is O(log(n)).

当前做法是O(NlogN)

如果我们频繁的操作该树，并且频繁的调用kth函数，有什么优化方法使时间复杂度降低至O(h)？h是树的高度。根据提示，我们可以在TreeNode中加入一个rank成员，这个变量记录的是该节点的左子树中节点的个数，其实就是有多少个节点比该节点小。这样我们就可以用二叉树搜索的方法来解决这个问题了。这个添加rank的操作可以在建树的时候一起完成。

 1   public int kthSmallest(TreeNode root, int k) {
 2         int count = countNodes(root.left);
 3         if (k <= count) {
 4             return kthSmallest(root.left, k);
 5         } else if (k > count + 1) {
 6             return kthSmallest(root.right, k-1-count); // 1 is counted as current node
 7         }
 8 
 9         return root.val;
10     }
11 
12     public int countNodes(TreeNode n) {
13         if (n == null) return 0;
14 
15         return 1 + countNodes(n.left) + countNodes(n.right);
16     }

Leetcode: Kth Smallest Element in a BST

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