669. 修剪二叉搜索树

• 刷题https://leetcode.cn/problems/trim-a-binary-search-tree/description/
• 文章讲解https://programmercarl.com/0669.%E4%BF%AE%E5%89%AA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html
• 视频讲解https://www.bilibili.com/video/BV17P41177ud/?share_source=copy_web&vd_source=af4853e80f89e28094a5fe1e220d9062

• 题解（递归法）：

/**
 * 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 {
    //递归
    public TreeNode trimBST(TreeNode root, int low, int high) {
        //若传入二叉树为空，则直接返回null
        if(root == null){
            return null;
        }
        //若当前结点小于规定区间，则当前结点的右分支还有可能在规定区间内
        //所以需要对当前结点右分支进行递归判断
        if(root.val < low){
            TreeNode right = trimBST(root.right, low, high);
            //将递归下一层得到的新的右分支根结点返回给当前层的父结点
            //当前root结点的删除
            return right;
        }
        //若当前结点大于规定区间，则当前结点的左分支还有可能在规定区间内
        //所以需要对当前结点左分支进行递归判断
        if(root.val > high){
            TreeNode left = trimBST(root.left, low,high);
            return left;
        }
        //若root在规定范围内
        //跟结点左分支递归
        root.left = trimBST(root.left, low, high);
        //根结点右分支递归
        root.right = trimBST(root.right, low, high);
        //返回新的二叉搜索树的根结点
        //前面也已经处理过只有根结点的情况，所以这里不用特殊处理根结点
        return root;
    }
}

108.将有序数组转换为二叉搜索树

• 刷题https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/description/
• 文章讲解https://programmercarl.com/0108.%E5%B0%86%E6%9C%89%E5%BA%8F%E6%95%B0%E7%BB%84%E8%BD%AC%E6%8D%A2%E4%B8%BA%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91.html
• 视频讲解https://www.bilibili.com/video/BV1uR4y1X7qL/?share_source=copy_web&vd_source=af4853e80f89e28094a5fe1e220d9062

• 题解（递归法）：

/**
 * 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 {
    //递归法(左闭右闭)
    public TreeNode sortedArrayToBST(int[] nums) {
        return sortedArrayToBST(nums, 0, nums.length - 1);
    }
    public TreeNode sortedArrayToBST(int[] nums, int left, int right){
        //因为左闭右闭，所以left = right 为合法区间
        //则left > right为非法区间
        if(left > right){
            return null;
        }
        //若只有一个结点，则直接充当根结点，返回
        if(right - left == 0){
            return new TreeNode(nums[left]);
        }
        //若有两个及以上结点，进行递归
        //注意左闭右闭的递归对应
        int mid =( left + right ) / 2;
        TreeNode root = new TreeNode(nums[mid]);
        //左递归
        root.left = sortedArrayToBST(nums, left, mid - 1);
        //右递归
        root.right = sortedArrayToBST(nums, mid + 1, right);
        //返回根结点
        return root;
    }
}

538.把二叉搜索树转换为累加树

• 刷题https://leetcode.cn/problems/convert-bst-to-greater-tree/description/
• 文章讲解https://programmercarl.com/0538.%E6%8A%8A%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E8%BD%AC%E6%8D%A2%E4%B8%BA%E7%B4%AF%E5%8A%A0%E6%A0%91.html
• 视频讲解https://www.bilibili.com/video/BV1d44y1f7wP/?share_source=copy_web&vd_source=af4853e80f89e28094a5fe1e220d9062

• 题解（递归法）：

/**
 * 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 {
    //累加
    //记录累加和
    int sum;
    public TreeNode convertBST(TreeNode root) {
        sum = 0;
        convertBST1(root);
        return root;
    }
    //由题意得，按右中左顺序遍历，进行累加即可
    public void convertBST1(TreeNode root){
        //若传入空树，则直接返回
        if(root == null){
            return;
        }
        //右处理
        convertBST1(root.right);
        //中处理
        //记录当前累加和
        sum += root.val;
        //将当前累加和赋值给当前结点
        root.val = sum;
        //左处理
        convertBST1(root.left);
    }
}