掌握二叉搜索树的双指针 + 公共祖先加深对后序遍历和递归的理解

530.二叉搜索树的最小绝对差

int min = Integer.MAX_VALUE;
    TreeNode pre;

    /**
     * <A href="https://leetcode.cn/problems/minimum-absolute-difference-in-bst/description/">530. 二叉搜索树的最小绝对差</A>
     */
    public int getMinimumDifference1(TreeNode root) {
        if (root == null) {
            return 0;
        }
        travel1(root);
        return min;
    }

    public void travel1(TreeNode cur) {
        if (cur == null) {
            return;
        }
        travel1(cur.left);
        if (pre != null) {
            min = Math.min(min, (cur.val - pre.val));
        }
        pre = cur;
        travel1(cur.right);
    }

501.二叉搜索树中的众数

    /**
     * <A href="https://leetcode.cn/problems/find-mode-in-binary-search-tree/">501. 二叉搜索树中的众数</A>
     */
    int ModeLength = 0;
    int count = 0;
    List<Integer> arr;
	TreeNode pre;
    public int[] findMode(TreeNode root) {
        arr = new ArrayList<>();
        travel2(root);
        int[] answer = new int[arr.size()];
        for (int i = 0; i < arr.size(); i++) {
            answer[i] = arr.get(i);
        }
        return answer;
    }

    public void travel2(TreeNode cur) {
        if (cur == null) {
            return;
        }
        travel2(cur.left);
        if (pre == null || pre.val != cur.val) {
            count = 1;
        } else {
            count++;
        }
        pre = cur;
        if (count > ModeLength) {
            arr.clear();
            ModeLength = count;
            arr.add(cur.val);
        } else if (count == ModeLength) {
            arr.add(cur.val);
        }
        travel2(cur.right);
    }

236. 二叉树的最近公共祖先

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }
        if(root.val == p.val || root.val == q.val){
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left,p,q);
        TreeNode right = lowestCommonAncestor(root.right,p,q);
        if(left  != null && right != null){
            return root;
        }else if(left==null){
            return right;
        }else{
            return left;
        }
    }