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?
思路， in order traversal，
注意第一个元素的推断， 设root为当前节点， 则 root.left == null && count == 0，就保证了root为 1st smallest element。样例：
4
/
/
2
/
/
1
4
/
/
2
3
上面两个bst中第一个节点分别为 1 和 2
/**
* 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> count = new ArrayList<Integer>();
count.add(0);
helper(root, count, k);
return count.get(1);
}
private void helper(TreeNode root, List<Integer> count, int target){
if(root == null){
return;
}
helper(root.left, count, target);
if(root.left == null && count.get(0) == 0){
count.set(0, 1);
}else{
count.set(0, count.get(0) + 1);
}
if(count.get(0) == target){
count.add(root.val);
return;
}
helper(root.right, count, target);
}
}