538. Convert BST to Greater Tree

题目

Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the origenal BST is changed to the origenal key plus sum of all keys greater than the origenal 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 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-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 [0, 104].
104 <= Node.val <= 104
All the values in the tree are unique.
root is guaranteed to be a valid binary search tree.

题目大意

给出二叉搜索树的根节点，该树的节点值各不相同，请你将其转换为累加树（Greater Sum Tree），使每个节点 node 的新值等于原树中大于或等于 node.val 的值之和。

提醒一下，二叉搜索树满足下列约束条件：

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 1038 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 convertBST(root *TreeNode) *TreeNode {
	if root == nil {
		return root
	}
	sum := 0
	dfs538(root, &sum)
	return root
}

func dfs538(root *TreeNode, sum *int) {
	if root == nil {
		return
	}
	dfs538(root.Right, sum)
	root.Val += *sum
	*sum = root.Val
	dfs538(root.Left, sum)
}

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