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897. Increasing Order Search Tree

Difficulty Topics

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:

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

Solution

increasing-order-search-tree.py
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def increasingBST(self, root: TreeNode) -> TreeNode:
        arr = []

        deq = collections.deque([root])

        while deq:
            node = deq.popleft()

            arr.append(node.val)

            for n in (node.left, node.right):
                if n:
                    deq.append(n)

        arr.sort()

        head = curr = TreeNode(arr[0])

        for num in arr[1:]:
            c = TreeNode(num)
            curr.right = c
            curr = curr.right


        return head
increasing-order-search-tree.cpp
/**
 * 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* increasingBST(TreeNode* root, TreeNode* tail = NULL) {
        if (!root) return tail;
        TreeNode* res = increasingBST(root->left, root);
        root->left = NULL;
        root->right = increasingBST(root->right, tail);
        return res;
    }
};