897. Increasing Order Search Tree (Python)

Related Topic

Depth-First-Search. Tree.

Description

Given 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 1 right child.

Sample I/O

Example 1

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \ 
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9  

Constraints:

• The number of nodes in the given tree will be between 1 and 100.
• Each node will have a unique integer value from 0 to 1000.

Methodology

Use DFS to append all tree node value into array in inorder. Then build a new tree with all nodes on the right.

Code

def increasingBST(self, root: TreeNode) -> TreeNode:
        if not root:
            return
        def inOrder(root, res):
            if not root:
                return
            inOrder(root.left, res)
            res.append(root.val)
            inOrder(root.right, res)
            return res
        array = inOrder(root, [])
        root = cur = TreeNode(array[0])
        for i in range(1, len(array)):
            cur.right = TreeNode(array[i])
            cur = cur.right
        return root

BigO

Time complexity: O(n) where n is the number of the nodes of both trees.