230.kth-smallest-element-in-a-bst

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230.kth-smallest-element-in-a-bst

230. 二叉搜索树中第K小的元素

给定一个二叉搜索树，编写一个函数 kthSmallest 来查找其中第 k 个最小的元素。

说明： 你可以假设 k 总是有效的，1 ≤ k ≤ 二叉搜索树元素个数。

示例 1:

输入: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
输出: 1

示例 2:

输入: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / \
     3   6
    / \
   2   4
  /
 1
输出: 3

进阶：

如果二叉搜索树经常被修改（插入/删除操作）并且你需要频繁地查找第 k 小的值，你将如何优化 kthSmallest 函数？

解法一

递归 中序遍历

/**
 * 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) {
        return inOrder(root).get(k - 1);
    }

    private List<Integer> inOrder(TreeNode root) {

        if (root == null) return new ArrayList<>();

        List<Integer> left = inOrder(root.left);
        List<Integer> right = inOrder(root.right);

        List<Integer> result = left;
        left.add(root.val);
        left.addAll(right);
        return result;
    }
}

解法二

借助 栈

/**
 * 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;

        Stack<TreeNode> stack = new Stack<>();

        while(true) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            if (--k == 0) return root.val;
            root = root.right;
        }
    }
}