#Leetcode# 897. Increasing Order Search Tree

https://leetcode.com/problems/increasing-order-search-tree/

Given a 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 1 right child.

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / 
    3    6
   /     
  2   4    8
 /        /  
1        7   9

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

 1
  
   2
    
     3
      
       4
        
         5
          
           6
            
             7
              
               8
                
                 9  

Note:

1. The number of nodes in the given tree will be between 1 and 100.
2. Each node will have a unique integer value from 0 to 1000.

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* increasingBST(TreeNode* root) {
       TreeNode* R = new TreeNode(-1);
        TreeNode* cur = R;

        inorder(root, cur);
        
        cur = R -> right;
        delete R;
        
        return cur;
    }
    void inorder(TreeNode* root, TreeNode* &cur) {
        if(root) {
            inorder(root -> left, cur);
        cur->right = root;
        root->left = NULL;
        cur = cur->right;
        inorder(root -> right, cur);
        }
        
    }
};

　　500！打卡