一、530.二叉搜索树的最小绝对差

题目链接：https://leetcode.cn/problems/minimum-absolute-difference-in-bst/
文章讲解：https://programmercarl.com/0530.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E7%9A%84%E6%9C%80%E5%B0%8F%E7%BB%9D%E5%AF%B9%E5%B7%AE.html
视频讲解：https://www.bilibili.com/video/BV1DD4y11779

1.1 初见思路

1. 二叉搜索树的定义：左子树的所有元素的值都小于中节点，右子树的所有元素的值都大于中节点
2. 二叉搜索树的特性：二叉搜索树的中序遍历后的数组是一个有序数组

1.2 具体实现

class Solution {TreeNode pre;int res = Integer.MAX_VALUE;public int getMinimumDifference(TreeNode root) {getMin(root);return res;}public void getMin(TreeNode node) {if (node == null) {return;}getMin(node.left);if (pre != null) {res = Math.min(res, Math.abs(node.val - pre.val));}pre = node;getMin(node.right);}
}

1.3 重难点

二、 501.二叉搜索树中的众数

题目链接：https://leetcode.cn/problems/find-mode-in-binary-search-tree/
文章讲解：https://programmercarl.com/0501.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E4%B8%AD%E7%9A%84%E4%BC%97%E6%95%B0.html
视频讲解：https://www.bilibili.com/video/BV1fD4y117gp

2.1 初见思路

1. 转换成有序数组，然后遍历一遍数组就可以得到结果
2. 可以在遍历树的时候就得到结果

2.2 具体实现

class Solution {int times = 0;int maxTimes = 0;TreeNode pre;List<Integer> res = new ArrayList<Integer>();public int[] findMode(TreeNode root) {findRes(root);int[] resArr = new int[res.size()];for (int i = 0; i < res.size(); i++) {resArr[i] = res.get(i);}return resArr;}// 中序遍历：左中右public void findRes(TreeNode node) {if (node == null) {return;}findRes(node.left);if (pre == null || pre.val != node.val) {times = 1;}if (pre != null && pre.val == node.val) {times++;}if (times > maxTimes) {maxTimes = times;res.clear();res.add(node.val);} else if (times == maxTimes) {res.add(node.val);}pre = node;findRes(node.right);}}

2.3 重难点

三、 236. 二叉树的最近公共祖先

题目链接：https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/
文章讲解：https://programmercarl.com/0236.%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88.html
视频讲解：https://www.bilibili.com/video/BV1jd4y1B7E2

3.1 初见思路

1. 无思路

3.2 具体实现

class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if(root==null || root==p || root==q){return root;}TreeNode left = lowestCommonAncestor(root.left,p,q);TreeNode right = lowestCommonAncestor(root.right,p,q);if(left==null){return right;}if(right==null){return left;}return root;}}

3.3 重难点