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:

https://assets.leetcode.com/uploads/2019/05/02/tree.png

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 #

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 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)
}

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