1038. Binary Search Tree to Greater Sum Tree

# 1038. Binary Search Tree to Greater Sum Tree#

## 题目 #

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 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.

Note: This question is the same as 538:  https://leetcode.com/problems/convert-bst-to-greater-tree/

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]
``````

Example 3:

``````Input: root = [1,0,2]
Output: [3,3,2]
``````

Example 4:

``````Input: root = [3,2,4,1]
Output: [7,9,4,10]
``````

Constraints:

• The number of nodes in the tree is in the range `[1, 100]`.
• `0 <= Node.val <= 100`
• All the values in the tree are unique.
• `root` is guaranteed to be a valid binary search tree.

## 题目大意 #

• 节点的左子树仅包含键 小于 节点键的节点。
• 节点的右子树仅包含键 大于 节点键的节点。
• 左右子树也必须是二叉搜索树。

## 解题思路 #

• 根据二叉搜索树的有序性，想要将其转换为累加树，只需按照 右节点 - 根节点 - 左节点的顺序遍历，并累加和即可。
• 此题同第 538 题。

## 代码 #

``````package leetcode

import (
"github.com/halfrost/leetcode-go/structures"
)

// TreeNode define
type TreeNode = structures.TreeNode

/**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/

func bstToGst(root *TreeNode) *TreeNode {
if root == nil {
return root
}
sum := 0
dfs1038(root, &sum)
return root
}

func dfs1038(root *TreeNode, sum *int) {
if root == nil {
return
}
dfs1038(root.Right, sum)
root.Val += *sum
*sum = root.Val
dfs1038(root.Left, sum)
}
``````

Apr 8, 2023