897. Increasing Order Search Tree

 897. 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

The solution to the above question is -

/**

 * 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* ans=NULL;

    void inorder(TreeNode* root){

        if(root==NULL)return;

        inorder(root->left);

        root->left=NULL;

        ans->right=root;

        ans=root;

        inorder(root->right);

        

    }

    

    TreeNode* increasingBST(TreeNode* root) {

        if(root==NULL)return root;

        TreeNode* res= new TreeNode(0);

        ans=res;

        inorder(root);

        return res->right;

    }

};