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.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
Output: 1
Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
Output: 3
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() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Comparator<Integer> comparator = new Comparator<Integer>() {
            @Override
            public int compare(Integer s1, Integer s2) {
                return s2 - s1;
            }
        };
        
        PriorityQueue<Integer> pq = new PriorityQueue<> (comparator);
        helper(root, k, pq);
        return pq.peek();
    }

    public void helper(TreeNode root, int k, PriorityQueue<Integer> pq){
        if (root == null) {
            return;
        }
        helper(root.left, k, pq);
        if (pq.size() < k) {
            pq.add(root.val);
        } 
        else {
            if (root.val < pq.peek()) {
                pq.poll();
                pq.add(root.val);
            }
        }
        helper(root.right, k, pq);
    }
}

标签：pq,TreeNode,val,BST,int,Smallest,Kth,right,root
来源： https://www.cnblogs.com/incrediblechangshuo/p/12826570.html