力扣hot100 - 230、二叉搜索树中第K小的元素

题目：思路：二叉树搜索树是按照左中右的顺序存储数据的，所以中序遍历刚好可以遍历出递增的数据。定义一个全局遍历count，和一个结果res存储结果，count初始化为k，每当遍历一个数字都count--，当count == 1时存入结果。

/** * 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 count = 0; public int res = 0; public int kthSmallest(TreeNode root, int k) { count = k; dfs(root); return res; } public void dfs(TreeNode root){ if(root == null){ return; } dfs(root.left); count--; if(count == 0){ res = root.val; return; } dfs(root.right); } }