538-convert-bst-to-greater-tree

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Description

Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.

As a reminder, a binary search tree is a tree that satisfies these constraints:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

 

Example 1:

Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]

Example 2:

Input: root = [0,null,1]
Output: [1,null,1]

 

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -104 <= Node.val <= 104
• All the values in the tree are unique.
• root is guaranteed to be a valid binary search tree.

 

Note: This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

Solution(Python)

# 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 __init__(self):
        self.total = 0

    def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        return self.reverseInorderMorris(root)

    def recursive(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        if root is not None:
            self.recursive(root.right)
            self.total += root.val
            root.val = self.total
            self.recursive(root.left)
        return root

    def reverseInorderMorris(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def get_succ(node):
            succ = node.right
            while succ.left is not None and succ.left is not node:
                succ = succ.left
            return succ

        total = 0
        node = root
        while node is not None:
            if node.right is None:
                total += node.val
                node.val = total
                node = node.left
            else:
                succ = get_succ(node)
                if succ.left is None:
                    succ.left = node
                    node = node.right
                else:
                    succ.left = None
                    total += node.val
                    node.val = total
                    node = node.left
        return root