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:
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)
}