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/**

* 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?

*

*/

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode(int x) { val = x; }

* }

*/

public class KthSmallestElementInABST230 {

public int kthSmallest(TreeNode root, int k) {

Stack<Integer> st = new Stack<>();

kthSmallest(root, k, st);

return st.pop();

}

public void kthSmallest(TreeNode root, int k, Stack<Integer> st) {

if (root == null) return;

kthSmallest(root.left, k, st);

if (st.size() == k) return;

st.push(root.val);

if (st.size() == k) return;

kthSmallest(root.right, k, st);

}

public int kthSmallest2(TreeNode root, int k) {

int[] i = new int[]{0, 0};

kthSmallest(root, k, i);

return i[1];

}

public void kthSmallest(TreeNode root, int k, int[] i) {

if (root.left != null) kthSmallest(root.left, k, i);

if (i[0] == k) return ;

i[0] = i[0] + 1;

i[1] = root.val;

if (i[0] == k) return;

if (root.right != null) kthSmallest(root.right, k, i);

}

public int kthSmallest3(TreeNode root, int k) {

int l = kth(root.left);

if (l >= k) {

return kthSmallest3(root.left, k);

} else if (l + 1 == k) {

return root.val;

} else {

return kthSmallest3(root.right, k-l-1);

}

}

public int kth(TreeNode root) {

if (root == null) return 0;

return kth(root.left) + kth(root.right) + 1;

}

}