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;
}
};
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