897. Increasing Order Search Tree

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897. Increasing Order Search Tree

Description

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:

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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:

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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

Hints/Notes

  • N/A

Solution

Language: C++

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/**
 * 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* tail = nullptr;

    TreeNode* increasingBST(TreeNode* root) {
        if (!root)
            return root;

        TreeNode* newHead = root;

        if (root->left) {
            newHead = increasingBST(root->left);
            root->left = nullptr;
        }

        if (tail)
            tail->right = root;
        tail = root;
        increasingBST(root->right);
        return newHead;
    }
};