文章目录

• 1. 题目
• 2. 思路
• • (1) 中序遍历+递归
  • (2) 中序遍历+栈
• 3. 代码

1. 题目

2. 思路

(1) 中序遍历+递归

• 中序遍历二叉搜索树，每遍历一个结点，k减1，当k为0时，该结点的值即为第k个最小元素。

(2) 中序遍历+栈

• 与(1)的思路基本相同，利用栈代替递归实现中序遍历。

3. 代码

import java.util.Deque;
import java.util.LinkedList;

public class Test {
    public static void main(String[] args) {
    }
}

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 {
    private int k = 0;
    private int res = -1;

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

    private void inorder(TreeNode root) {
        if (root == null || res >= 0) {
            return;
        }
        inorder(root.left);
        k--;
        if (k == 0) {
            res = root.val;
        }
        inorder(root.right);
    }
}

class Solution1 {
    public int kthSmallest(TreeNode root, int k) {
        Deque<TreeNode> stack = new LinkedList<>();
        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            k--;
            if (k == 0) {
                break;
            }
            root = root.right;
        }
        return root.val;
    }
}