題目

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.

題目大意

給出二叉 搜索 樹的根節點，該樹的節點值各不相同，請你將其轉換為累加樹（Greater Sum Tree），使每個節點 node 的新值等於原樹中大於或等於 node.val 的值之和。

提醒一下，二叉搜索樹滿足下列約束條件：

• 節點的左子樹僅包含鍵 小於 節點鍵的節點。
• 節點的右子樹僅包含鍵 大於 節點鍵的節點。
• 左右子樹也必須是二叉搜索樹。

解題思路

• 根據二叉搜索樹的有序性，想要將其轉換為累加樹，只需按照 右節點 - 根節點 - 左節點的順序遍歷，並累加和即可。
• 此題同第 538 題。

參考代碼

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)
}