Problem Description

Given the root of a binary search tree with distinct values, modify it so that every node has a new value equal to the sum of the values of the original tree that are greater than or equal to node.val.

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: [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]

Note:

The number of nodes in the tree is between 1 and 100. Each node will have value between 0 and 100. The given tree is a binary search tree.

Solution

Since the given tree is binary search tree, we can do reverse inorder traversal of the tree and keep the sum of all nodes that have been visited and update current node val to the sum. We’ve modified the tree in place. Below is python implementation

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def bstToGst(self, root: TreeNode) -> TreeNode:
        self.sum = 0
        self.dfs(root)
        return root
        
        
    def dfs(self, root: TreeNode)->None:
        if root == None:
            return
        self.dfs(root.right)
        self.sum += root.val
        root.val = self.sum
        self.dfs(root.left)
        return