897. Increasing Order Search Tree

 897Increasing 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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