算法|修剪二叉搜索树、将有序数组转换为二叉搜索树、把二叉搜索树转

一、 669. 修剪二叉搜索树

题目连接：https://leetcode.cn/problems/trim-a-binary-search-tree/
思路：遍历二叉树，比最小是还小的 向root.righ遍历，比最大值还大的root.left遍历，符合条件的正常遍历

/**
 * 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 TreeNode trimBST(TreeNode root, int low, int high) {
        if (root == null) return null;
        //比最小值还小，需要裁减
        if (root.val < low) return trimBST(root.right, low, high);
        //比最大值还打，需要裁减
        if (root.val > high) return trimBST(root.left, low, high);
        root.left = trimBST(root.left, low, high);
        root.right = trimBST(root.right, low, high);
        return root;
    }
}

二、 108. 将有序数组转换为二叉搜索树

题目连接：https://leetcode.cn/problems/convert-sorted-array-to-binary-search-tree/

/**
 * 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 {
    private TreeNode sortedArrayToBST(int[] nums, int left, int right) {
        if (left >= right) return null;
        int mid = left + ((right - left) / 2);
        TreeNode treeNode = new TreeNode(nums[mid]);
        treeNode.left = sortedArrayToBST(nums, left, mid);
        treeNode.right = sortedArrayToBST(nums, mid + 1, right);
        return treeNode;
    }
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0) return null;
        return sortedArrayToBST(nums, 0, nums.length); 
    }
}

三、 538. 把二叉搜索树转换为累加树

题目连接：https://leetcode.cn/problems/convert-bst-to-greater-tree/
思路：中序遍历，先遍历右节点，记录sum为累加，sum += root.val root.val = sum;

/**
 * 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 {
    private int sum = 0;
    private void convert(TreeNode root) {
        if (root == null) return;
        convert(root.right);
        sum += root.val;
        root.val = sum;
        convert(root.left);
    }
    public TreeNode convertBST(TreeNode root) {
        convert(root);
        return root;
    }
}