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106 lines
2.7 KiB
Markdown
106 lines
2.7 KiB
Markdown
# [538. Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/)
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## 题目
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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.
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As a reminder, a *binary search tree* is a tree that satisfies these constraints:
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- The left subtree of a node contains only nodes with keys **less than** the node's key.
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- The right subtree of a node contains only nodes with keys **greater than** the node's key.
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- Both the left and right subtrees must also be binary search trees.
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**Note:** This question is the same as 1038: [https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/](https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/)
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**Example 1:**
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```
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Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
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Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
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```
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**Example 2:**
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```
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Input: root = [0,null,1]
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Output: [1,null,1]
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```
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**Example 3:**
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```
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Input: root = [1,0,2]
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Output: [3,3,2]
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```
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**Example 4:**
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```
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Input: root = [3,2,4,1]
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Output: [7,9,4,10]
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```
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**Constraints:**
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- The number of nodes in the tree is in the range `[0, 104]`.
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- `104 <= Node.val <= 104`
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- All the values in the tree are **unique**.
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- `root` is guaranteed to be a valid binary search tree.
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## 题目大意
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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。
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提醒一下，二叉搜索树满足下列约束条件：
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- 节点的左子树仅包含键 小于 节点键的节点。
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- 节点的右子树仅包含键 大于 节点键的节点。
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- 左右子树也必须是二叉搜索树。
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## 解题思路
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- 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。
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- 此题同第 1038 题。
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## 代码
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```go
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package leetcode
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import (
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"github.com/halfrost/LeetCode-Go/structures"
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)
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// TreeNode define
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type TreeNode = structures.TreeNode
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func convertBST(root *TreeNode) *TreeNode {
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if root == nil {
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return root
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}
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sum := 0
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dfs538(root, &sum)
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return root
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}
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func dfs538(root *TreeNode, sum *int) {
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if root == nil {
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return
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}
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dfs538(root.Right, sum)
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root.Val += *sum
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*sum = root.Val
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dfs538(root.Left, sum)
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}
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``` |