LC.P230[二叉搜索树中第K小的元素]

方法一：遍历+优先队列

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/**
 * 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> q = new PriorityQueue<>();
        preOrder(root, q);
        while (--k > 0) q.poll();
        return q.peek();
    }

    private void preOrder(TreeNode root, PriorityQueue<Integer> q) {
        if (root == null) return;
        q.offer(root.val);
        preOrder(root.left, q);
        preOrder(root.right, q);
    }
}
  • 时间复杂度：$O(nlogn)$
  • 空间复杂度：$O(n)$

方法二：中序遍历

利用二叉搜索树的中序遍历为递增序列这一特点，对树进行中序遍历计数。

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/**
 * 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 {
    int ans, k;

    public int kthSmallest(TreeNode root, int k) {
        this.k = k;
        inOrder(root);
        return ans;
    }

    private void inOrder(TreeNode root) {
        if (root == null) return;
        inOrder(root.left);
        if (k == 0) return;
        if (--k == 0) ans = root.val;
        inOrder(root.right);
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$