230. Kth Smallest Element in a BST

Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.

Example 1:

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

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3

Solution:

/**
 * 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) {
        if (root == null){
            return -1;
        }

        PriorityQueue<TreeNode> maxHeap = new PriorityQueue<TreeNode>(new Comparator<TreeNode>(){
            @Override
            public int compare(TreeNode t1, TreeNode t2){
                if (t1.val > t2.val){
                    return -1;
                }else if (t1.val == t2.val){
                    return 0;
                }else{
                    return 1;
                }
            }
        });

        DFS(root, maxHeap, k);
        return maxHeap.poll().val;

    }

    private void DFS(TreeNode root, PriorityQueue<TreeNode> maxHeap, int k){
        if (root == null){
            return;
        }

        if (maxHeap.size() < k){
            maxHeap.offer(root);
        }else if (maxHeap.peek().val > root.val){
            maxHeap.poll();
            maxHeap.offer(root);
        }

        DFS(root.left, maxHeap, k);
        DFS(root.right, maxHeap, k);
    }
}
//TC: O(n)
//SC: O(n)

/**
 * 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) {
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>(Collections.reverseOrder()); // max heap 初始化为最大堆
        helper(root, k, pq);
        return pq.poll(); 
    }

    private void helper(TreeNode root, int k, PriorityQueue<Integer> pq){
        if (root == null){
            return;
        }

      if (pq.size() < k){ // If size is not k yet then add any element
            pq.add(root.val);
        }else if (root.val < pq.peek()){
            pq.poll();  // pop the top
            pq.add(root.val); // add cur value to pq
        }

        helper(root.left, k, pq);
        helper(root.right, k, pq);
        return;
    }
}

//Time complexity: O(n) + O(log(n)) = O(n)
//Space complexity: O(n)