Problem
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 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/
Example 1:
image
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.
-
rootis guaranteed to be a valid binary search tree.
问题
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。
注意：本题和 1038: https://leetcode-cn.com/problems/binary-search-tree-to-greater-sum-tree/ 相同
示例 1：
image
输入：[4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
输出：[30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
示例 2：
输入：root = [0,null,1]
输出：[1,null,1]
示例 3：
输入：root = [1,0,2]
输出：[3,3,2]
示例 4：
输入：root = [3,2,4,1]
输出：[7,9,4,10]
提示：
- 树中的节点数介于 0 和 104 之间。
- 每个节点的值介于 -104 和 104 之间。
- 树中的所有值 互不相同 。
- 给定的树为二叉搜索树。
思路
中序遍历
利用 BST 的中序遍历就是升序的特性，降序遍历 BST 的元素值。
Python3 代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def convertBST(self, root: TreeNode) -> TreeNode:
def dfs(root):
nonlocal sumval
if root:
dfs(root.right)
sumval += root.val
root.val = sumval # 将BST转化成累加树
dfs(root.left)
sumval = 0
dfs(root)
return root
