LeetCode-897. Increasing Order Search TreeLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/increasing-order-search-tree/

Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.

Example 1:

Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

Example 2:

Input: root = [5,1,7]
Output: [1,null,5,null,7]

Constraints:

• The number of nodes in the given tree will be in the range [1, 100].
• 0 <= Node.val <= 1000

【C++】

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* increasingBST(TreeNode* root) {
        vector<int> vals;
        inorder(root, vals);
        TreeNode* ans = new TreeNode(0), *cur = ans;
        for (int v: vals) {
            cur->right = new TreeNode(v);
            cur = cur->right;
        }
        return ans->right;
    }
 
    void inorder(TreeNode* node, vector<int>& vals) {
        if (node == nullptr) return;
        inorder(node->left, vals);
        vals.push_back(node->val);
        inorder(node->right, vals);
    }
};

【Java】

/**
 * 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 increasingBST(TreeNode root) {
        List vals = new ArrayList();
        inorder(root, vals);
        TreeNode ans = new TreeNode(0), cur = ans;
        for (int v: vals) {
            cur.right = new TreeNode(v);
            cur = cur.right;
        }
        return ans.right;
    }
 
    public void inorder(TreeNode node, List vals) {
        if (node == null) return;
        inorder(node.left, vals);
        vals.add(node.val);
        inorder(node.right, vals);
    }
}